Download The free energy is a measure of the amount of mechanical (or other

The Free Energy
The free energy is a measure of the amount of mechanical (or other)
work that can be extracted from a system, and is helpful in engineering
applications. It is a linear combination of the energy and the entropy of a
system, yielding a thermodynamic state function which represents the
"useful energy". It uses a mathematical "trick" to produce a function which
automatically accommodates any entropy change due to heat exchanged
with the surroundings.
Definition of the Helmholtz energy
The Helmholtz energy is defined as:
A = U - TS
(1)
where




A is the Helmholtz free energy,
U is the internal energy of the system,
T is the absolute temperature,
S is the entropy.
From the first law of thermodynamics we have:
with:
and:
dU = δq + δw
δw = δw* -PdV
δq = TdS
(w* is the useful work)
dU = TdS –PdV + δw*
dA = TdS –PdV + δw* - SdT – TdS
dA = -PdV - SdT + δw*
(2)
We can see that at constant T and V we have:
(dA)T,V = δw*
or (∆A)T,V = w*
1
For process without useful work (w* = 0):
dA = PdV– SdT
(3)
The Gibbs free energy
The Gibbs free energy is a thermodynamic potential which measures
the "useful" work obtainable from an isothermal, isobaric thermodynamic
system. Technically, the Gibbs free energy is the maximum amount of nonpV work which can be extracted from a closed system, and this maximum
can only be attained in a completely reversible process. When a system
evolves from a well-defined initial state to a well-defined final state, the
Gibbs free energy ΔG equals the work exchanged by the system with its
surroundings, less the work of the pressure forces, during a reversible
transformation of the system from the same initial state to the same final
state.
Definitions
The Gibbs free energy is defined as:
G = H – TS = U + PV – TS





where:
U is the internal energy
P is pressure
V is volume
T is the temperature
S is the entropy
dG = dU + PdV + VdP – TdS – SdT
dG = δq + δw + PdV + VdP – TdS – SdT
but we have : δw = δw* -PdV and : δq = TdS
where w* is the useful work.
dG = TdS + δw* -PdV + PdV + VdP – TdS – SdT
dG = VdP– SdT + δw*
(4)
We can see that at constant T and P we have:
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(dG)T,P = δw*
or (∆G)T,P = w*
For process without useful work (w* = 0):
dG = VdP – SdT
(5)
For isothermal process of an ideal gas:
P2
P2
nRT
P
dP  nRTLn 2
P
P1
P1
dG = VdP  G   VdP  
P1
(6)
The Maxwell relations:
Starting from the equation (5) and using the properties of state function of G
G
G
dG = VdP– SdT =   dP    dT
 P T
 T  P
G
G
we obtain :    V and    S
 P T
 T  P
and differentiating second time we obtain:
 S 
 V 
   

 P T
 T  P
(7)
In a similar manner, and starting from the following equations:
dA = PdV– SdT
dU = TdS - PdV
dH = - PdV– SdT
we will find
 T 
 P 
 T   V 
 P   S 

    ,    
 ,   

 V T
 S V
 P  S  V  P  T V  V T
(7bis)
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Free energy changes for chemical reactions
Usually the reactions are carried out isothermally, then we have for
any reaction:
RP
∆A = ∆U - T∆S = ∆U – T (SP – SR)
(7)
∆G = ∆H - T∆S = ∆H – T (SP – SR)
(7bis)
Example:
The standard heat of formation of NH3 is -46.11 kJ mol-1 and the
standard entropies of N2 , H2 , and NH3 are respectively 191.6, 130.68, and
192.45 J K-1 mol-1
Calculate the variation of Gibbs energy for the reaction:
N2 + 3 H2  2 NH3
Answer:
∆G = ∆H - T∆S
∆S = 2 S(NH3) – S(N2) – 3 S(H2) = - 198.74 J K-1
∆H = 2 ∆Hf(NH3) = - 92.22 kJ mol-1
∆G = - 92220 – 298(-198.74) = 32995.48 J
The spontaneous reaction
The Gibbs energy is more common in chemistry than Helmholtz
function, because we are usually interested in changes occurring at constant
pressure, not constant volume. Then, the criterion for spontaneous reaction
is related to Gibbs energy:
(dG)T,P < 0 or (∆G)T,P < 0
which means that :
“chemical reactions are spontaneous in the direction of
decreasing Gibbs energy”
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By using equation (7bis) , this criterion can be expressed as
function of ∆H and ∆S signs:
∆H
+
+
∆S
+
+
∆G
+
+,+,-
The reaction is
spontaneous
not spontaneous
spontaneous at low temperature
spontaneous at high temperature
N.B. “IF ∆G = 0, then the reaction is in equilibrium.”
Example:
For the reaction seen above:
N2 + 3 H2  2 NH3
Which temperatures will favor the formation of NH3?
Answer:
We have found that:
∆S = - 198.74 J K-1
∆H = - 92.22 kJ mol-1
∆G = 32995.48 J (the reaction is not spontaneous at 298 K)
and because ∆S < 0 ; ∆H < 0 , the reaction could be spontaneous at low
temperature.
Standard Gibbs free energy
Definition: the Standard Gibbs free energy, Go, of a substance at some
temperature T is equal to its Gibbs free energy at 1 atm pressure.
From equation (6) we get:
G  G (T , P2 )  G (T , P1 )  nRTLn
P2
P1
Then we choose P1 = 1 atm, and we obtain:
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G  G(T , P2 )  G o (T )  nRTLn ( P2 )
and for any temperature T and pressure P:
G(T , P)  G o (T )  nRTLn ( P)
(8)
here, in equation (8), P must be in atm.
Standard change of formation
The standard Gibbs free energy of formation of a compound is the change
of Gibbs free energy that accompanies the formation of 1 mole of that
substance from its component elements, at their standard states (the most
stable form of the element at 25 degrees Celsius and 100 kilopascals). Its
symbol is ΔGfO.
All elements in their standard states (oxygen gas, graphite, etc.) have 0
standard Gibbs free energy change of formation, as there is no change
involved.
The standard Gibbs free energy variation for some chemical reaction
aA + bB  cC +dD
is given by
0
Greaction
  G 0f (Pr oducts)   G 0f (Re ac tan ts)
(9)
 cG 0f (C )  dG 0f ( D)  aG 0f ( A)  bG 0f ( B)
Chemical Potential
Definition: the Chemical Potential is equal to the Molar Gibbs free energy.
 (T , P) 
G (T , P)
n
(9)
Combining (8) and (9):
 (T , P)   o (T )  RTLn ( P)
(10)
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Phases’ equilibrium
Phase(1) ↔ Phase(2)
At equilibrium
(∆G)T,P = 0 = G2 – G1 => G2 = G1
A new equilibrium after small change in T and P
G2 + dG2 = G1 + dG1 => dG2 = dG1
-S2dT + V2dP = -S1dT + V1dP
dP S 2  S1 S
H



dT V2  V1 V TV
(11)
The above equation is said “Clapeyron Equation” and can be integrated as
follow:
 First case, small change in T :
P H

T TV
(12)
 Second case, important change in T :
P2
H
P dP  V
1
T2
dT
T
T1
 P2  P1 
H  T2 
ln  
V  T1 
(13)
Example: What is the boiling temperature of water under 770 torr pressure?
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Free energy and the equilibrium constant
For a gaseous reversible reaction:
aA + bB ↔ cC +dD
we have:
Kp 
PCc PDd
PAa PBb
(14)
and:
 ( A)   o ( A)  RTLn ( PA )
 ( B)   o ( B)  RTLn ( PB )
 (C )   o (C )  RTLn ( PC )
 ( D)   o ( D)  RTLn ( PD )
the free energy variation for the reaction is:
G   G(Pr oducts)   G(Re ac tan ts)
 cG(C )  dG( D)  aG( A)  bG( B)
and by using molar free energy
G  cCo  d Do  a Ao  b Bo  RT cLnPC  dLnPD  aLnPA  bLnPB 
 G o  RTLn
PCc PDd
 G o  RTLn ( K p )
a b
PA PB
G  Go  RTLn ( K p )
……. (15)
At equilibrium
G  G o  RTLn ( K p )  0
G o   RTLn ( K p )
  G 0 

K p  exp 
 RT 
……. (16)
Other forms of equilibrium constant
By using the ideal gas relation :
PiV = niRT (i means A, B, C, or D)
which gives:
Pi = [i]RT
and combining this last relation with eq. (14):
K p  Kc RT 
n g
(17)
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with
Kc 
C c D d
Aa B b
In the same manner we can find that
K p  KxP
Kx 
with
n g
(18)
xCc xDd
x Aa xBb
where x is the molar fraction ( xi 
(19)
ni
)
 ni
Variation of the equilibrium constant with temperature
Starting from eq. (16):
 G 0
RT
 LnK p 
1   G o 
H o

 

  
R T  T  p RT 2
 T  p
Ln( K p ) 
Integrating between T1 and T2:
 K p  
H o  1 1 
T2

  
Ln 
R  T2 T1 
 K p T1 
(20)
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