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Chemical Thermodynamics
2013/2014
7th Lecture: Gibbs Energy and Fundamental Equations
Valentim M B Nunes, UD de Engenharia
What have we learned (from the 2nd Law!)
We have seen that, in isolated systems, the entropy change give us the
direction of an spontaneous transformation:
ΔStotal > 0 – spontaneous transformation
ΔStotal = 0 – equilibrium (reversible process)
But, systems of interest generally are not isolated. It will be
convenient to have a criteria for spontaneous change depending only on
the system.
Gibbs (1839-1903) introduced a new state function, that plays a key
role in chemistry and in the study of chemical equilibrium:
G  H  TS
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Josiah Willard Gibbs
Thermodynamicist!
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Criteria for Spontaneous Change
Consider a system in thermal equilibrium with its surroundings:
Tsyst  Tsurr
dSUniv  dS syst  dS surr
dSUniv  dS syst 
dQsurr
dQ
 dS syst  surr
Tsurr
Tsyst
dSUniv  dS syst 
dQsyst
dS syst 
dQsyst
By the second Law,
Tsyst
Tsyst
0
For any transformation in a closed system at constant p (p=pext)
dH
dS 
 0 or TdS  dH
T
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Criteria for Spontaneous Change
Under constant T= Tsurr and constant p = pext, the criterion for
spontaneity is:
dG  0
This means that at constant p and constant T, equilibrium is achieved
when the Gibbs energy is minimized.
Consider the process A  B (keeping p and T constant)
ΔG < O – A  B is spontaneous
ΔG = 0 – A and B are in equilibrium
ΔG > 0 – then B  A is spontaneous
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What tell us G? (some mistakes!)
G  H  TS
minimum
maximum
Wrong!
The second law is still valid. In some
transformations the maximization of
entropy can be obtained by
“transferring” enthalpy to the
surroundings
Sext  
Snowing: gas  liquid  solid
H syst
Text
If Text is low (cold air masses) then
|ΔSext| > | ΔSsyst|
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The Helmholtz energy
If we define now the Helmholtz energy, A, as:
A  U  TS
The criterion for spontaneity under constant T = Tsurr and constant
volume, V it will be:
dA  0
Gibbs energy is, by far, more important that Helmholtz energy.
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Physical meaning of Gibbs energy
At constant temperature:
dG  dH  TdS
Since H = U + pV
dG  dU  pdV  Vdp  TdS  dQ  dW  Vdp  pdV  TdS
dG  dW  Vdp  pdV
Now, considering that the total work is the sum of expansion work and
other types of work
dW   pdV  dWmáx, except expansion
As result, at constant p we finally obtain
dG  dWmáx
The Gibbs energy change represents the maximum available work, or
the useful work that we can obtain from a given process.
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Standard Gibbs energy
As for enthalpy we can also obtain standard Gibbs energy of formation,
ΔGºf (at pº = 1 bar). For any element in their reference state,
ΔGºf=0.
From the definition of G = H – TS we obtain:
Gº  H ºTS º
and:
Gr0   i G 0f ,i
i
These values are tabled, and are obtained from
calorimetric or spectroscopic data!
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The Gibbs Energy and Fundamental Equations
We can now combine the first an second law; as we have introduced all
the state functions for closed systems
From the 1st Law:
dU  dQ  dW
then:
dU  TdS  pdV
U= U(S,V)
Since H = U + pV
G = H-TS
A=U-TS
dH  TdS  Vdp
H = H(S,p)
dG  Vdp  SdT
G = G(p,T)
dA   pdV  SdT
A =A(T,V)
FUNDAMENTAL
EQUATIONS:
valid for
reversible or
irreversible
changes for
closed systems
and only pV
work.
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How G changes with p and T
Since G is a state function, and as we saw before G = G(T,p) then
we can write:
 G 
 G 
 dp  
dG  
 dT
 T  p
 p T
And easily obtain:
 G 

  S
 T  p
and
 G 

  V
 p T
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How G changes with p and T
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The Maxwell Relations
Recalling that df = gdx + hdy is an exact differential if
and using the fundamental equations we obtain
 S 
 V 






 T  p
 p T
 T 
 p 






 V  S
 S V
 g   h 
    
 y  x  x  y
Maxwell Relations: these
equations allow us to establish
useful thermodynamic relations
between state properties!
 p 
 S 





 T V  V T
 T

 p

 V 
  

 S  S  p
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U and H from equations of state
We can now relate U and H to p-V-T data. For instance for U, we
can calculate the ΠT value:
 U   U   S   U 
 
 
 

 V T  S V  V T  V  S
T  
 p 
T  T    p
 T V
For an ideal gas:
Then:
nR p
 p 


 
 T V V T
 U 
T  
 0
 V T
Generic equation of
state!
This proves that for an
ideal gas U = U(T), that
is function of T only !
RT
a
 U 
For a van der Waals gas: 


p

0

2
Vm
 V T Vm  b
U=U(T,V)
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The variation of Gibbs Energy with temperature
As we saw previously:
 G 

  S

T

p
then:
GH
 G 



T
 T  p
If we derivate now G/T we obtain:

T
or:
And finally:

 G  1  G 


G
 


T
T

T
T
 


 1  1  G  G
  
 2
 T  T  T  T
 G 1 GH  G
  
 2
T  T  T  T  T
 
 G
T

 T

H
 2
T

p
Gibbs-Helmholtz equation
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Gibbs Helmholtz equation and Chemical Reactions
Applying the Gibbs Helmholtz equation to a chemical reaction we
obtain:

  G
T

 T


H
 2
T

p
Where ΔH is the reaction enthalpy change.
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The variation of Gibbs Energy with pressure
Recalling that
 G 

  V
 p T
And upon integration we obtain, at constant T:
 dG   Vdp
pf
G  G  p f   G  pi    V ( p)dp
pi
For liquids and solids the molar volume, Vm, is small and approximately
constant, so:
G  p f   G  pi   Vm p
~0
G  p f   G  pi 
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Ideal Gases
For a perfect gas (pV = nRT) we obtain:
or:
G  p f   G  pi   nRT
pf

pi
G  p f   G pi   nRT ln
1
dp
p
pf
pi
Considering now that pi = pº (the standard pressure of 1 bar) and
calculating the molar Gibbs energy, Gm (n=1), we obtain:
Gm ( p)  Gm0  RT ln
p
po
The molar Gibbs energy is also called the chemical potential, μ:
p
    RT ln 0
p
o
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Final remark
We have now all the “thermodynamic tools” that allow us to apply the
thermodynamic principles to all kind of physical and chemical processes,
like chemical equilibrium or phase equilibrium. That will be the syllabus
for the next lectures…
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