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Equilibrium and Kinetics
Chapter 2
Recap
In the last lecture we used the mechanical
Analogy to understand the concept of
Stability and metastability
Recap
Fig. 2.2
Mechanical push to overcome
activation barrier
unstable
Activation
barrier
P.E
metastable
System automatically
attains the stable state
stable
Configuration
Recap
If we want to transform the Local Minimum
- METASTABLE to Global Minimum - Most
STABLE then we have to overcome the
activation barrier (could be by mechanical
push, thermal activation)
Thermodynamic functions
U = internal
energy
At constant pressure
Courtsey: H. Bhadhesia
t

This expression can also be expressed as: U = Uo + Cv dt
Courtsey: H. Bhadhesia
o
Sum of internal energy and external energy
For solids and liquid the PV term is negligible
t
The above expression can also be expressed as: H = Ho +
Courtsey: H. Bhadhesia
 C dt
p
o
Courtsey: H. Bhadhesia
P
Courtsey: H. Bhadhesia
Entropy
Courtsey: H. Bhadhesia
How do you measure the entropy?
Courtsey: H. Bhadhesia
Gibbs Free Energy
G  H  TS
(2.6)
Condition for equilibrium
≡ minimization of G
Local minimum ≡ metastable equilibrium
Global minimum ≡ stable equilibrium
G = GfinalGinitial
(2.7)
G = 0 
reversible change
G < 0 
irreversible or
(2.8)
spontaneous change
G > 0 
impossible
The variation of G with temperature
Atomic
or
statistical
interpretation of entropy
Entropy
The entropy of a system can be defined by
two components:
Thermal:
Configurational:
S  k ln W
Boltzmann’s
Epitaph
S  k ln W
(2.5)
W is the number of
microstates
corresponding to a given
macrostate
N!
W  Cn 
n!( N  n)!
N
N=16, n=8, W=12,870
(2.9)
If n>>>1
Stirling’s Approximation
ln n! n ln n  n
(2.11)
S  k ln W
N!
 k ln
(2.10)
n!( N  n)!
 k [ N ln N  n ln n  ( N  n) ln( N  n)]
(2.12)
KINETICS: Arrhenius equation
Svante Augustus
Arrhenius
 Q 
rate  A exp  

 RT 
1859-1927
(2.15)
Nobel
1903
Rate of a chemical reaction varies with temperature
Arrhenius plot
 Q 
rate  A exp  

 RT 
ln (rate)
slope  
Fig. 2.4
Q
R
1
T
Thermal energy
Average thermal energy per atom
per mode of oscillation is kT
Average thermal energy per mole
of atoms per mode of oscillation is
NkT=RT
(2.13)
Maxwell-Boltzmann Distribution
n
 E 
 exp  

N
 kT 
(2.14)
Fraction of atoms having an energy  E
at temperature T
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