# Entropy change in reversible process by malj

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```									Entropy change in reversible
process
Entropy change in reversible process
At constant temp. (Isothermal )

At constant volume (Isochoric )

At constant pressure(Isoparic)
Entropy of mixing
1- Before
mixing
1 mole A
+
1 mole B

(mixed)

Initial state

To calculate the summation of entropy before mixing for ni where ni = n1+ n2
Entropy of mixing
After mixing

1 mole B
Total entropy change

+
Example
• Exactly one liter of 0.1M solution of a
substance A is added to 3 liters of a
0.05M solution of a substance B.
Assume ideal behavior and calculate
the entropy of mixing.
• Answer . 0.1 mol of substance A is present and
the volume increases by a factor of 4:
• ∆ S(A)= 0.1x8.314xln 4/1
•        = 1.153 J/K
• 0.15 mol of B is present and the volume increases
by a factor 4/3
• ∆ S(B)= 0.15x8.314x ln (4/3)
•        = 0.359 J/ K
• ∆ S(total)= 1.153 + 0.359 = 1.512 J/K
Questions
• 1-An amount of gas 0.5 mol allowed to
expand against constant pressure at 300K
from 1 L to 10 L ,Calculate the changes of the
entropy.
• 2- An ideal gas at initial pressure of 20 atm
and 300 K occupies a volume of 1 L .calculate
the entropy of the system for reversible
isothermal expansion of the gas to a final
volume 10 L.
Questions
• 3-Calculate the entropy changes when one
mole of an ideal gas expand reversibly from
one liter to 10 L at 25 C.
• 4-Evaluate the change in entropy when 3
moles of a gas are heated from 27C at a
constant pressure of 1 atm. Where the heat
capacity of the gas is 23.7 J/ K.
Second Law of Thermodynamics
The second law of thermodynamics: The
entropy of the universe does not change for
reversible processes and increases for
spontaneous processes.
Reversible (ideal):

Irreversible (real, spontaneous):
• Three possibilities:
– If DSuniv > 0…..process is spontaneous
– If DSuniv < 0….(non spontaneous and process
is spontaneous in opposite direction.
– If DSuniv = 0….equilibrium

•   Here’s the catch: We need to know DS for both
the system and surroundings to predict if a
reaction will be spontaneous!
Combined first and second law of thermodynamics
•
2nd law
Internal energy and Entropy
Helmohpltz Free Energy
A = U –TS

Enthalpy and entropy
• Gibbs Free Energy

G =H –TS
Entropy and Phase Changes

• Phase Change: Reaction in which a substance
goes from one phase of state to another.

• Example:
H2O(l)      H2O(g) @ 373 K

• Phase changes are equilibrium processes such
that:
DSuniv = 0
Phase changes
• As an example of phase changes ,the melting
point of a solid. At melting point, the liquid
and solid exist together in equilibrium. The
addition of a little heat would liquefy some of
the solid, but the solid liquid equilibrium will
be maintained. This process is reversible and
entropy change can be given as
• Phase changes

The temp. is constant
Example
• Determine the temperature at which liquid bromine boils :
Br2(l)       Br2(g)
• where DH°f(Br2(g)) =30.91 kJ
S°(Br2 (g)) 245.38 J/K
S°(Br2(l))= 152.23 J/K
0
DH° = DH°f(Br2(g)) - DH°f(Br2(l))
= 30.91 kJ - 0
= 30.91 kJ
Now, DS° = S°(Br2 (g)) - S°(Br2(l))
= 245.38 J/K - 152.23 J/K
= 93.2 J/K
DS°sys = 93.2 J/K = DHsys/T
Questions
• 1- Calculate the entropy changes involved in
the conversion of 1 mole of ice at 25C to liquid
at 1 atm; the enthalpy of fusion per mole of
ice is 6008 J/mole.
• 2- Calculate the entropy change for a process
the following process.
• a) 1 mole H2o(l, 100°C,1 atm )→1 mole H2o(v,
100°C,1 atm ) given that
•    ∆Hvap = 40850 J /mol
Questions
If the enthalpy of vaporization of water
∆Hvap(H2O) = 9717 cal/mol at normal
boiling point ,calculate the entropy
accompanying the vaporization of 2
moles of H2O at 100 °C and 1 atm.
The Third Law

• The third law: The entropy of a perfect crystal at
0K is zero.

• The third law provides the reference state for use
in calculating absolute entropies.
What is a Perfect Crystal?

Perfect crystal at 0 K   Crystal deforms at T > 0 K
Standard Entropies

• Recall, entropy is a state function;
therefore, the entropy change for a
chemical reaction can be calculated as
follows:

DS o
reaction    S
o
prod .    So
react.
Example

• Balance the following reaction and
determine DS°.
Fe(s) + H2O(g)         Fe2O3(s) + H2(g)
2Fe(s) + 3H2O(g)         Fe2O3(s) + 3H2(g)
DS°= (S°(Fe2O3(s)) + 3S°H2(g))
-(2S°Fe(s) + 3S°H2O(g))

DS°= -141.5 J/K
Question
Calculate the standard entropy change for the
following reactions
H2(g)+ Cl2(g) → 2 HCl(g) ,
standard entropy HCl =186.68 ,Cl2= 222.95 ,
H2= 130.59 JK-1mol-1

2 H2O(l) → 2 H2(g)+ O2(g)         ,S° H2O =130.6
,O2=205 JK-1mol-1
Effect of temperature on standard
entropy
• For a chemical reaction at constant pressure
Example
• Calculate the change of standard entropy at
333K for the following reaction

Substance     S°298 J mol-1K-1   Cp JK-1
C2H5OHl           160.5            111.4
HIg              206.3            29.13
C2H5Il            211.5           114.95
H2Ol               69.8           75.22
•   Helmholtz free energy
•   A=U-TS
•   (dA)T =dWnet
•   The change of Helmholtz free energy is equal
to the maximum reversible work which a
system can do isothermally.

• For isothermal process
•   Gibss free energy
•   G=H-TS
•   (dG)T,P =dWnon-PV
•   The change of Gibss free energy is equal to
the non-PV work

• For isothermal process
Maxwell’s Relations
• From Helmholtz
• A=U-TS

• From Gibss
Dependence of Gibss Free energy on
temperature at constant pressure

Gibbs –Helmholtz equation
problem
• 1-An ideal gas at 300K expand isothermally
from 10 to 1 atm. Against a pressure that
gradually reduced ,calculate q,w, DU , DH, DG
, DA, DS.
• 2- Four moles of an ideal gas are compressed
isothermally at 300 K from 2atm to 4 atm.
Calculate the free energy change for the
process.
Criteria for spontaneous equilibrium
change
• Condition of equilibrium.
• Irreversible process is accompanied by a net
increase of entropy of the system and it’s
surrounding.

• For reversible process
Criteria for spontaneous equilibrium
change(Free energy)
• (spontaneous process)
• (dA)T,V= negative value irreversible process
• Non spontaneous process
• (dA)T,V= 0              reversible process
• Non spontaneous and spontaneous in reverse
direction
• (dA)T,V= positive value
•
Criteria for spontaneous equilibrium
change
• (spontaneous process)
• (dG)T,p= negative value irreversible process
• Non spontaneous process
• (dG)T,p= 0              reversible process
• Non spontaneous and spontaneous in reverse
direction
• (dG)T,p= positive value
Criteria for spontaneous equilibrium
change
• Spontaneous process accompanied by a
decrease of work function
• Helmholtz free energy decreases in natural
process when it reaches minimum value the
system reaches equilibrium
Gibbs Free Energy
1. If DG is negative, the
forward reaction is
spontaneous.
2. If DG is 0, the system is at
equilibrium.
3. If DG is positive, the
reaction is spontaneous in
the reverse direction.
Standard Free Energy Changes
Standard free energies of formation, DGf are
analogous to standard enthalpies of
formation, DHf.

DG can be looked up in tables,
or
calculated from S° and DH.
Standard Free Energy of formation
• The change of free energy due to the
formation of one mole of compound from it’s
• element in a standard state .
• DGf for any element in standard state equal
zero.
Free Energy Changes

Very key equation:

∆G = ∆H-T∆S

This equation shows how DG changes with
temperature.
(We assume S & DH are independent of T.)
Free Energy and Temperature

• There are two parts to the free energy
equation:
 DH— the enthalpy term
 TDS — the entropy term

• The temperature dependence of free
energy comes from the entropy term.
Free Energy and Temperature

By knowing the sign (+ or -) of DS and DH,
we can get the sign of DG and determine if a
reaction is spontaneous.
The Contribution to Spontaneous Change

ΔG = ΔH – TΔS

Enthalpy Change        Entropy Change   Spontaneous?

ΔH < 0 (Exothermic)       ΔS > 0        Yes (ΔG < 0)
ΔH < 0 (Exothermic)       ΔS < 0        Yes, if |TΔS| < |ΔH|
ΔH < 0 (Exothermic)       T=0           Yes (ΔG < 0)
ΔH > 0 (Endothermic)      ΔS > 0        Yes, if |TΔS| > |ΔH|
ΔH > 0 (Endothermic)      ΔS < 0        No (ΔG > 0)
ΔH > 0 (Endothermic)      T=0           No (ΔG > 0)

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Example
For the following reaction
NH4NO3(s) + 3 H2(g)→ 3H2O(g) + N2H4(s)
Substance            ∆H°f k J mol -      S°J K-1 mol-1
NH4NO3               -365                     150
H2(g)                 0.0                      130
H2O(g)              -242                        139
N2H4                 50                         120
Calculate ∆ G°
R= 0.082 L atm K-1mol-1
= 8.314 J K-1mol-1
= 1.987 cal K-1mol-1
Cal = 4.18 J
0.082 L atm K-1mol-1 = 8.314 J K-1mol-1 = 1.987 cal K-1mol-1
1 L atm =101.3 J
cal = 0.041 L atm
Physical equilibria involving phase
transition Clapeyron Clausius equation
• In system of pure substance if several phases are
present at equilibrium under a given set of
conditions, there is always a possibility of transition
of substance from one to another by altering any of
the variables of the system.
• A thermodynamic relation between the change of
pressure with change of temperature of a system at
equilibrium is called clapeyron equation.
Physical equilibria involving phase transition
Clapeyron Clausius equation
Application of Clapeyron equation
• 1- liquid –Vapour equilibria (vaporization
process)

Ex. The heat of vaporization of H2O is 40820 Jmol-1,the molar
volume of liquid H2O is 18.78 ml, and the molar volume of steam is
30.199 L all at 100°C and 1 atm . what would be the change in
boiling point of H2O at 100°C if the atmospheric pressure is
changed by 1mm (1L atm =101.3 J)
Application of Clapeyron equation
• 2- Solid –liquid equilibria (fusion process)

Ex. Calculate the change of atmospheric pressure to
cause a change of one degree in melting point of
benzene . Data given tfus=5.0°C heat of fusion =128.6
J/g. The volume per gm of liquid and solid benzene
are 1.119 and 1.06 cm3 respectively.
Application of Clapeyron equation
• Calculate the change in the freezing point of water at
0°C per one atm change of pressure . Where, at 0°C
,heat of fusion of ice is 335J/g. The density of water
is 0.9998g/ cm3 and density of ice is 0.9168
Application of Clapeyron equation
• 3- Solid-Vapour equilibria (Sublimation)

4-Equilibrium between two crystalline forms

Calculate the rate of change of transition temperature with
pressure for sulpher. Where transition temp. =95.5°C at 1
atm ,enthalpy of transition per gm of sulpher is 13.4 J.
Monoclinic sulpher has a greater specific volume than
that of rhombic sulpher by 0.0126cm3 /g.
Clausis-Clapeyron Equation
• When applied to the liquid vapour equilibria,
with the assumption that the volume of the
liquid phase is negligible as compared to that
of vapour phase which behave as a perfect
gas. The clausis –Clapeyron equation used, if
the temperature is not near the critical point,
then,
Clausis-Clapeyron Equation
• Vg ˃˃ VL   , ∆V= Vg , Vg =RT/P
Example
• The melting point of a solid at 3 atmosphere is
107°C,the change in volume during fusion is
15 C.C g-1 and ∆Hf=513 cal/mol. Find the
melting point of this solid if the pressure is
raised to 730 atm.
The standard Gibbs free energy (G°)
• Standard free energy of a substance at one
atm
• SG°= f(T) or G = G°(T)
• ∆ G =nRT lnP2 / P1
• G(T,P) - G°(T) =nRTln P
• G(T,P) =G°(T) +nRTln P
Free Energy and Equilibrium
Let us consider the reaction
aA +bB         mM + nN

Variation of equilibrium constant and temperature
( Vant Hoff equation.
Problem
• Calculate the free energy change when one
mole of a gas change it’s pressure at 298K
from( i) p=100 atm to 1 atm.
• (ii)p = 1 atm to p = 100 atm
• State the type of the process.
Problem
•   Calculate the values of Kp at 25°C and 800°C for the following reaction
•   Using the following data,
•   Substance
•   ∆G°(kJ)               0            -137.27        -228.59         -394.38
•   ∆H°(kJ)              0            -110.52        -241.83           -392.51
INTRODUCTION TO STATISTICAL
THERMODYNAMICS
• A. Basic Concepts and Postulates
• Statistical thermodynamics is the branch of
chemistry and physics that seeks to calculate
the thermodynamic behavior of macroscopic
systems starting from molecular properties;

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• properties of individual molecules or small
groups of molecules
• Try to predict
• equilibrium thermodynamic properties of
macroscopic systems

58
– Thermodynamics
• study of the relationships between
macroscopic properties
–Volume, pressure, compressibility, …
– Statistical Mechanics (Statistical
Thermodynamics)
• how the various macroscopic properties
arise as a consequence of the microscopic
nature of the system
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Position and momenta of individual
molecules (mechanical variables)
– Statistical Thermodynamics (or Statistical
Mechanics) is a link between microscopic
properties and bulk (macroscopic) properties
• Description of States
– Macrostates : T, P, V, … (fewer variables)
– Microstates : position, momentum of each
particles (~1023 variables)
•
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• Related with the system microcosmic
and macrocosmic properties, it is
satisfied for some results we get from
the simple molecule. No need to
carry out the complicated low
temperature measured heat
experiment, then we can get the
quite exact entropy.
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• The structure model must be supposed when
calculating, certain approximate properties
exist; for large complicated molecules and
the agglomerated system, it still has some
difficulties in calculating.

62
• The use of statistical methods is necessary
because detailed application of either
quantum or classical mechanics to real
macroscopic systems is clearly impossible.

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Energy
Levels

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Configuration ....
At any instance, there may be no
molecules at e0 , n1 molecules at e1 ,
n2 molecules at e2 , …
 {n0 , n1 , n2 …}
configuration

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Configuration ....
At any instance, there may be no molecules at e0 , n1
molecules at e1 , n2 molecules at e2 , …
 {n0 , n1 , n2 …} configuration

e5

e4
e3
{ 3,2,2,1,0,0}
e2
e1
e0

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Weight ....
Each configurations can be achieved in
different ways
Example1 : {3,0} configuration  1

Example2 : {2,1} configuration  3

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Dominating Configuration
5                      5                       5
4                      4                       4
3                      3                       3
2                      2                       2
1                      1                       1
0                      0                       0

W = 1 (5!/5!)         W = 20 (5!/3!)           W = 5 (5!/4!)

Difference in W becomes larger when N is increased !
In molecular systems (N~1023) considering the
most dominant configuration is enough for average
W is thermodynamic probability

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