Entropy change in reversible process

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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
• 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.
Answer
                 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)




                                                               41
                      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;



                                                  57
• i.e., Start with
• 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
                                                 59
        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)
•
                                                  60
             Advantage
• 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.
                                        61
              Disadvantage
• 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.




                                                63
Energy
Levels




         64
Configuration ....
At any instance, there may be no
molecules at e0 , n1 molecules at e1 ,
n2 molecules at e2 , …
          {n0 , n1 , n2 …}
configuration


                                         65
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


                                                                       66
Weight ....
  Each configurations can be achieved in
  different ways
Example1 : {3,0} configuration  1

   Example2 : {2,1} configuration  3




                                        67
        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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