Thermodynamic Terminology PDF

11/10/2024 aA  bB cC  dD  ΔH  ( )p  Δ CpG.R kirchoff s equation T HT  HT  Cp (T2  T1) 2 1...

11/10/2024 aA  bB cC  dD  ΔH  ( )p  Δ CpG.R kirchoff s equation T HT  HT  Cp (T2  T1) 2 1 In case if Cp for reactants and products vary with temp for example Cp     T   T 2 ......... C p  (    T   T 2 .......) 1 1 ΔHT2  ΔHT1  Δα(T2  T1 )  Δβ(T22  T12 )  Δγ(T23  T13 ) ... 2 3 BOND ENERGY ΔH= bond formation+ Bond breaking Bond formation - Ve , Bond Braking + ve Second Law Of Thermodynamics Spontaneous Changes heat flows from a hotter body to a colder body, and chemical reactions proceed to equilibrium spontaneous changes do not reverse themselves 1  12/17/2024 Third Law of Thermodynamics What is a Perfect Crystal? Perfect crystal at 0 K Crystal deforms at T > 0 K 273 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero kelvin is 0. The third law used in calculating absolute entropies. S = kB ln Ω Where S is entropy, k is the Boltzmann constant = 1.380649×10−23 J K-1 and Ω is the number of different ways in which the energy of the system can be achieved by rearranging the atoms or molecules among their available states. ‫عدد الطرق المختلفة التي يمكن من خﻼلها تحقيق طاقة النظام عن طريق إعادة ترتيب الذرات أو الجزيئات بين حاﻻتها‬.‫المتاحة‬ 1 274  12/17/2024 The absolute value of entropy is known at T= 0K ( S is zero ). ⇒ the absolute value of entropy can be determined at any temperature Statistical explanation to the 3rd Law At temperature T = 0 K all particles of a perfect crystal are at the lowest energy level: this is possible in a single way: Ω = 1 ⇒ ln 1 = 0, S = k ln Ω ⇒ S = 0 J/K 275 Mathematical Explanation of the Third Law As statistical mechanics, the entropy of a system can be expressed via the following equation: S – S0 = 𝑘B ln𝛀 Where,  S is the entropy of the system.  S0 is the initial entropy. Now, for a perfect crystal that has exactly one unique ground state, 𝛀 = 1. Therefore, the equation can be rewritten as follows S – S0 = 𝑘B ln(1) = 0 [because ln(1) = 0] When the initial entropy of the system is selected as zero, the following value of ‘S’ can be obtained: S – S0 = 0 ⇒ S = 0 Thus, the entropy of a perfect crystal at absolute zero is zero. 2 276  12/17/2024 Applications of the Third Law of Thermodynamics An important application of the third law of thermodynamics is that it helps in the calculation of the absolute entropy of a substance at any temperature ‘T’. These determinations are based on the heat capacity measurements of the substance. For any solid, let S0 be the entropy at 0 K and S be the entropy at T K, then 𝑻𝑪 𝒑 ∆𝑺 = 𝑺 − 𝑺𝟎 = 𝒅𝑻 𝟎 𝑻 According to the third law of thermodynamics, S0= 0 at 0 K, 𝐶 𝑆= 𝑑𝑇 𝑇 277 The simplified expression for the absolute entropy of a solid at temperature T is as follows: 𝑻𝑪 𝑻 𝒑 𝑺= 𝒅𝑻 = 𝑪𝒑 𝒍𝒏 𝑻 𝟎 𝑻 𝟎 S  CP ln T Here Cp is the heat capacity of the substance at constant pressure and this value is assumed to be constant in the range of 0 to T K. 3 278  12/17/2024 Thermodynamic Terminology Types of the system System properties THERMODYNAMIC EQUILIBRIUM THERMODYNAMIC PROCESSES  2Z 2Z Z = f (x , y)  yx xy  Z   Z  dZ    dx    dy   x y  y x 𝝏𝒛 𝝏𝒚 𝝏𝒙 ( )𝒙 ( )𝒛 ( )𝒚 = −𝟏 cyclic rule 𝝏𝒚 𝝏𝒙 𝝏𝒛 279 qv= E qP= H First Law of Thermodynamics E = q - w E = q - PdV Compression in a single step. Compression in more steps. W = PV W = P(Vf – Vi) W = L. atm 1 L atm = 101.3 J Expansion in a single step. Expansion in more steps. 4 280  12/17/2024 H   E    V    E  C    C    p   T p C p  C v    + P     V T   T  P v  T v Work done by free expansion = 0 CP – CV = R for ideal gad  E  Joule experiment   0  V T Work isothermal reversible process V2 P1 E= H = 0 q  w  n R T ln q  w  n R T ln V1 P2 Work in irreversible isothermal P2 w = q = nRT(1- ) E= H = 0 P1 281 in reversible adiabatic Decrease of internal energy results in the lowering of temperature in case of adiabatic q=0 E = - w expansion increase of temperature for adiabatic E = nCV (T2 – T1) = -w compression w= nCV (T1 – T2) H = nCp (T2 – T1) R /C T  V  V 1 2 1    T2  V1  T 1  V   2    T1  V2  T2 V2R / Cv  T1V1 1     T  P  2   2  R /C P T2 P 1     2  1  CP / CV  P2  V1       P1  V 2  PV 1 1 PV 2 2 T  P  T1  P1  1  1  C v  ( P 2 / P1 ) R Adiabatic Irreversible Expansion T 2  T1 Cp 5 282  12/17/2024 Comparison between isothermal and adiabatic process dP P Lower than slope in adiabatic   γ dV V dP P slope in isothermal  - dV V E = q- w The process is (vapour) If the system absorbs heat q is positive. (liquid) If heat lost q is negative w= PV = P(Vv – VL) if value of Vv and VL is If work is done by the system, is positive. given but if not given Work done on the system is negative. = PVv if Vv > VL then neglect VL w = RT 283 Thermochemistry aA+bB cC+ dD HReaction =  n (Hf )P -  n(Hf )R H = E + ngas RT  ΔH  ( )p  Δ CpG.R kirchoff s equation H - E = ngas RT T qp = qv + ngas RT Hess’s Law H  C p T HT  HT  Cp (T2  T1) 2 1 1 1 ΔH T2  ΔH T1  Δα(T 2  T1 )  Δβ(T 22  T12 )  Δγ(T 23  T13 ) ... 2 3 H = H1 + H2 + H3 + H4 BOND ENERGY Bond formation - Ve , ΔH =  bond formation +  Bond breaking Bond Braking + ve 6 284  12/17/2024 Second Law Of Thermodynamics dq rev Spontaneous Changes  T 0 Carnot Heat Engine 2 dq S  rev S2  S1 dS  dq rev Total work w cyc q  q1 T T    2 1 Heat absorbed q 2 q2 mathematical statements of the second law of T2  T1 T q   1 1 1 1 thermodynamics. T2 T2 q2 0 < < 1 285 V  P  reversible adiabatic ΔS  nRln  2  reversible ΔS  nR ln  1  processes are called  V1  Isothermal  T2  isentropic processes. Srev (System) = Sirr (System) (state function) But q rev not equal q irr Entropy Change on Heating qirr (sys) = nRT ( 1 – P2/P1) = - qsurr or Cooling of a Substance S surr = - q/T system in irriversible process Reversible Adiabatic Changes S  0 T  ΔS  nC V ln  2  ΔH trans  T1  ΔS trans  T  Phase changes. Ttrans ΔS  nC P ln  2   T1  Srev (Sys) = -Srev (surr) ΔSuniverse  ΔSsystem  ΔSsurroundings ΔSuniverse  ΔSsystem  ΔSsurroundings  0 Entropy Change for Isolated System (dS)rev = 0 and (dS)irr > 0 Or (S)rev = 0 and (S)irr > 0 7 286  12/17/2024 Combined Form of the First and Second Laws of thermo-dynamics dE = TdS – PdV  T P Maxwell equation(1)      VS SV dH = TdS + VdP T V Maxwel equation (2)     PS  SP A = E - TS 1-Helmholtz free energy A -(A)T = wrev P   S      Maxwell`s equation (3) dA= -PdV – SdT  T  v  V  T V  P   A  E A  nRT ln  2    nRT ln 1     2   V1  P2 T  T v T 287 2- Gibbs free energy G G = H - TS - (G)T,P = wnet  S   V       Maxwell's (4) dG = -SdT+VdP  P  T  T  P P2 V   (G / T )   H  G  nRT ln  nRT ln 1    P1 V2 T P T2 Application of the Gibbs-Helmholtz Equation G  H  T S Zn | Zn SO4 (soln)// CuSO4 (soln)| Cu  E o   E o  -G = nF Eo  nFE  H  nFT o  S  nF    T P  T P 8 288  12/17/2024 (A)system (G)system ΔSuniverse  ΔSuniverse T T Since for a spontaneous process the entropy change of the universe must be positive, (Asystem , Gsystem) must be negative. (i) Criteria for equilibria at constant T and V. ( ΔA) T, V  0 (ΔG) T,P  0 for a reversible process ( A)T , V 0 In an irreversible process (spontaneous changes) ( ΔA) T, V  0 (ii) Criteria for equilibria at constant T and P. for a reversible process (G)T , P  0 For irreversible (spontaneous) changes (ΔG)T , P  0 289 Property Sign Nature of Process (G)T,P or (A)T,V Negative Spontaneous Zero Equilibrium Positive Non-Spontaneous dP Sβ  S ΔS Clapeyron dP H dP   equation  HT V dT Vβ  Vα ΔV dT T  V dT H T ( P2  P1 )  ln 2 V T1 This form of equation is usually employed for processes involving condensed phases. 9 290  12/17/2024 dP ΔH vap dP ΔH fus dP  H sub    dT T(V V  VL ) dT T(V L  VS ) dT T  V dP Δ Htrans  The Clausius- Clapeyron Equation dT T (V  V ) dP ΔHvap P ΔHvap ΔT RT 2    dT TVV RT2 ΔP ΔH vap P  H vap ln P    C a plot of ln P versus 1/T should be linear with a slope of Hvap / R RT P2 Δ H vap  T2  T1  H vap 1 1 ln ( )   ln P2  ln P1   [  ] P1 R  T1T2  R T2 T1 291 G = -R T ln K Chemical Equilibrium  K 2 H  T2  T1 d ln K H van't Hoff equation. ln  ( )  K1 R T1 T2 dT RT 2 ΔH ο So that plot of ln K against 1/T should be a straight line of ln K    constant slope -H/ R. RT 10 292  12/10/2024 Application of the Gibbs-Helmholtz Equation Zn | Zn SO4 (soln)// CuSO4 (soln)| Cu -G = nF Eo  E o   E o   nFE  H  nFT o  S  nF   T P   T  P 249 (A)system ΔSuniverse  T (G)system ΔSuniverse T Since for a spontaneous process the entropy change of the universe must be positive, (Asystem , Gsystem) must be negative. ( ΔA) T, V  0 (ΔG) T,P  0 1 250  12/10/2024 (i) Criteria for equilibria at constant T and V. for a reversible process ( A)T , V 0 In an irreversible process (spontaneous changes) ( ΔA) T, V  0 (ii) Criteria for equilibria at constant T and P. for a reversible process (G)T , P  0 For irreversible (spontaneous) changes (ΔG)T , P  0 251 Property Sign Nature of Process (G)T,P or (A)T,V Negative Spontaneous Zero Equilibrium Positive Spontaneous in reverse direction dP Sβ  S ΔS Clapeyron dP H dP   equation  HT V dT Vβ  Vα ΔV dT T  V dT H T ( P2  P1 )  ln 2 V T1 This form of equation is usually employed for processes involving condensed phases. 2 252  12/10/2024 dP ΔH vap dP ΔH fus dP  H sub    dT T(V V  VL ) dT T(V L  VS ) dT T  V dP Δ Htrans  The Clausius- Clapeyron Equation dT T (V  V ) dP ΔHvap P ΔHvap ΔT RT 2    dT TVV RT2 ΔP ΔH vap P  H vap ln P    C RT H vap 1 1 P2 Δ H vap  T2  T1  ln P2  ln P1   [  ] ln ( )   R T2 T1 P1 R  T1T2  253 Example The heat of vaporization of water is 40820 Jmole-1, the molar volume of liquid water and of steam are 16.78 ml and 30.199 liters all at 100 oC. What would be the change in the boiling point of water at 100 oC if the atmospheric is changed by 1 mmHg. ( 1liter atm = 101.3 J ) ΔH dP vap 40820    3.62 JK 1L1 dT T  VV  VL  373*  30.199  0.01678  dP 3.62   0.0357 atm deg 1  0.0357*760  27.19 mmHg deg 1 dT 101.3 dT 1   0.0368 deg/mmHg dP 27.19 3 254  12/10/2024 Example An organic compound X melts at 80oC. If the vapor pressure of the liquid X is 10mmHg at 85.8oC and 40 mm at 119.3oC , Calculate Heat of vaporization of the liquid, the boiling point and the entropy of vaporization at the boiling point P2 ΔHvap  T2  T1  ln ( )   P1 R  T1T2  40 ΔHvap  392.3  358.8  ΔHvap  48475J / mole  48.47kJ / mole ln ( )  358.8 *392.3  10 8.314 At the boiling point Tb the vapor pressure (P2) is one atm or 760mm Hg. P2 Hvap 1 1 760 48475 1 1 Tb = 489 K ln  [  ] ln  [  ] P1 R T2 T1 10 8.314 Tb 358.81 ΔSvap = ΔHvap /Tb =48475/489 =99.13 J/deg/mole 255 Chemical Equilibrium In the study of chemical processes generally two questions are importance. First is that how fast the reaction proceeds to form the product and the second question is that how far the reaction will proceed in a given direction? The answer of the first question constitutes the domain of the chemical kinetics and second answer is provided by the chemical equilibria. 4 256  12/10/2024 In this Chapter we shall consider the thermodynamic aspects of the chemical equilibria. Thermodynamic Derivation of the Law of Mass Action: The law of mass action as such cannot be derived ‫ﻻداﻋﻰ‬thermodynamically, what we can do is that we can deduce an expression for the equilibrium constant by thermodynamic method. This method for deriving the expression of equilibrium constant has the great merit of not requiring any postulated reaction mechanism. 257 Let us consider the general reaction Reactants = products a A + bB = c D+ dD in which the reactants and the products are ideal gases. The free energy of any mixture can be calculated from the following equation. 5 258  12/10/2024 dP  dG nRT P G = n RT ln p + constant Constant = G (the standard free energy)  G = G  + nRT ln p the free energy of any mixture of a mole of A and b moles of B is given as follows. 259 GA  GA  a RT ln PA GB  GB  b RT ln PB Similarly the free energy of the products is given as, GC  GC  c RT ln PC GD  GD  d RT ln PD 6 260  12/10/2024 GReaction = GP - GR     GοC  cRTlnPC  GοD  dRTln PD   G  aRTln P   G ο A A ο B  bRTln P B at equilibrium there is no free energy change G= 0 and equation reduces to,     GοC  cRTln PC  GοD  dRTln PD   G  aRTln P   G ο A A ο B   dRTln PB  0 261 G  G   G  G  RT ο c ο D ο A ο B  ln PCc  RT lnP Dd  RT ln P  RT ln P   0 a A b B PCc PDd ΔGο  RT ln 0 PAa PBb Pcc PDd or ΔGο   RT ln Pcc PDd PAa PBb  a b  K eq = equilibrium constant PA PB G = -R T ln K (75) 7 262  12/10/2024 Variation of Equilibrium Constant with Temperature : G = -R T ln K ΔGο ln K   d ln K 1 d (ΔGο/ T) RT . dT R dT (G/T) H   2  T P T d ln K H   dT RT 2 This equation is van't Hoff equation. 263 Integration of Van't Hoff Equation: If the change of heat content H were independent of temperature, general integration of the van't Hoff equation would give. d ln K H  H   dT RT 2  d ln K   RT 2 dT ΔHο ln K    constant RT 8 264  12/10/2024 So that plot of ln K against 1/T should be a straight line of slope -H/ R. Integration between the temperature limits of T1 and T2, the corresponding values of the equilibrium constant being K1 and K2, gives K 2 H  T2  T1 ln  ( ) K1 R T1 T2 265 Free Energy and Temperature By knowing the sign (+ or -) of S and H, we can get the sign of G and determine if a reaction is spontaneous. 9 266  12/10/2024 Example One mole of a perfect gas at 27oC expands isothermally and reversibly from10 atm to 1 atm against a pressure that is gradually reduced. Calculate q, w, and each of the thermodynamic quantities ΔE, ΔH, ΔG, ΔA, and ΔS. P  W  RTln  1  w  10   1.987*300.15*ln    1373cal mole 1 P  max  2  1 ΔE = 0 ΔE = q-w ΔH = ΔE + Δ(PV) =0 +0 =0 q = w =1373 cal mole-1 q 1373 ΔS rev  4.57 cal K1mole1 ΔA = -Wmax = -1373 cal mole-1 T 300 ΔHΔG 0(1373) P  1 alsoΔS  4.57 cal K1mole1 G  nRTln  2   1.987 * 300.15ln   1373 cal mole  1 T 300 P   10   1 267 Example For the synthesis of ammonia, Kp = 1.64*10-4 at 673 K when the reaction is represented by N2 (g) +3H2 (g) = 2NH3 (g) Calculate ΔGo for the reaction at the given temperature. ΔG o   RTlnK P ΔG o  8.314*673*ln(1.64*104 ) ΔG o  48775.6J  48.842 kJ 10 268  12/10/2024 Calculate ΔGo and the equilibrium constant for the following reaction at 25oC C  2H (g)  CH (g) (graphite) 2 4 using the given data: Standard enthalpy of the reaction ΔHo298 = -74830 J Sographite = 5.68 JK-1mole-1, SoHydrogen = 130.59 JK-1mole-1 , Sometane = 186.19 JK-1mole-1 ΔSo  So  2*So  So methane hydrogen graphite ΔSo  186.19  2*130.59  5.68  80.67 JK 1 ΔG o  ΔH o  TΔSo ΔG o  74830  298*(  80.67)  50790 J 50790 ΔG o   RTlnK  8.314*298*lnK lnK   20.4999 P P P 8.314*298 K  7.97*108 P 269 Example Calculate the value of Kp at 25oC and 800oC for the water gas reaction using the following data: CO (g)  H 2 O (g)  H 2 (g)  CO 2 (g) Substance H2 (g) CO (g) H2O (g) CO2 (g) ΔGof (KJ) 0.0 -137.27 -228.59 -394.38 ΔHof (KJ) 0.0 -110.52 -241.83 -392.51  Products   ΔGo reactants ΔG o   ΔG o ΔG o  (ΔG o (H )  ΔG o (CO )   (ΔG o (CO)  ΔG o (H O)   f 2 f 2   f f 2  ΔG o  0  (  394.38)  (  137.27)  (  228.59)  28.52 KJ     11 270  12/10/2024 ΔG o   RTlnK P 28.52 lnK   11.511 8.314*103*298 P K  9.98*104 At 25oC P ΔHo   ΔHo  P ro d u c t s    Δ H o re a c ta n t s Δ H o   ( Δ H o ( H )  Δ H o ( C O )    (Δ H o (C O )  Δ H o ( H O )   f 2 f 2   f f 2  Δ H o   0  (  3 9 2.5 1 )    (  1 1 0.5 2 )  (  2 4 1.8 3 )    4 0.1 6 k J K 2 H  T2  T1 ln  ( ) K1 R T1 T2  K  -4 0.1 6 1073  298  ln  2    9.9 8 * 1 0 4  8. 3 1 4 * 1 0  3  1 0 7 3 * 2 9 8    K  0.8 2 a t 8 0 0 o C 2 271 12  12/4/2024 Combined Form of the First and Second Laws of thermo-dynamics dE = TdS – PdV  T P Maxwell equation(1)      VS SV dH = TdS + VdP T V     Maxwel equation (2) PS  SP 1-Helmholtz free energy A A = E - TS -(A)T = wrev  P   S  dA= -PdV – SdT     Maxwell`s equation (3)  T  v  V  T 214 V  P   A  E A  nRT ln  2    nRT ln 1     2  V1  P2 T  T  v T 2- Gibbs free energy G G = H - TS - (G)T,P = wnet dG = -SdT+VdP  S   V        P  T  T  P Maxwell's (4) P2 V   (G / T )   H  G  nRT ln  nRT ln 1  P1 V2  T  P T2 1 215  12/4/2024 Application of the Gibbs-Helmholtz Equation Relation between electrical and chemical energy. Consider Daniell`s cell Zn | Zn SO4 (soln) // CuSO4 (soln)| Cu 216 If the e.m.f. of the reversible cell is Eo volts and the quantity of electricity passing is n Faradays (or nF Coulomb) then the electrical work done by the system is equal to n x F x Eo joules. Hence -G = nF Eo This relationship may be substituted in the Gibbs-Helmholtz equation to give  E o   E o  S  nF    nFE o  H  nFT   T P  T P 2 217  12/4/2024 Here (Eo/T)P is the rate of change of e.m.f. with temperature. Example For Daniell cell calculate G if Eo = 1.10 volt at 25o C and n=2; F = 96500 Coulomb Solution. For this case have G = -nF E = -2x (96500 Coulomb) x 1.10 Volt = -212300 Volt Coulomb ( Joule) 218 Example: For the Weston standard cell calculate G, H and S using the following data: Eo = 1.01463 Volts at 25o C; n=2; F=96500 (dEo /dT)P = -5.0 x 10-5 Volts/deg Solution. G = -nFEo  E o  = -195815 Joules  nFE o   H  nFT     T P -195815   H  2 * 96500 * 298 * -5.0 x 10 -5  E  o S  nF   -195815   H  2875.7  T P ΔH =-198690.7 Joule  S  2 * 9 6 5 0 0 * -5.0 x 1 0 -5  S = - 9.6 5 J 3 219  12/4/2024 Free Energy of the System and Entropy of Universe: (1) (A) System and (S) Universe Consider an isolated system consisting of system and the surroundings (thermostat) as shown in Fig. (a). spontaneous changes (a) at System is a vessel with a fixed volume and constant containing a substance which can undergo a temperature and chemical reaction volume (A)T,V < 0 The surrounding which is a thermostat is considered to have a large heat capacity so that any heat given to it or taken from it by the system will not change the bath temperature 220 Suppose that the contents of the vessel (system) undergo the volume of the vessel is constant no work is done by the system and w=0. first law for this process gives Eq where q is the heat absorbed by the system from the surroundings. Ssurrounding = -(qsystem/T)= - (Esystem/T) where T is the temperature of the thermostat 4 221  12/4/2024 Suniverse = Ssystem + Sunrroundings ΔEsystem  ΔSsystem T  1 T   ΔEsystem  T ΔSsystem  A= E - TS A= E - TS at constant T (A) system Suniverse   T 222 This proves that the change in the Helmholtz free energy of the system accounts automatically for the entropy changes of the system and surroundings. Since for spontaneous the entropy change of the universe must be positive, then (A) for the system must be negative. 5 223  12/4/2024 (2) (G) System and (S) Universe (G)system ΔSuniverse T 224 For spontaneous change at constant T and P (G)system 0 6 225  12/4/2024 Free Energy of the System and Entropy of Universe: (1) (A) System and (S) Universe  (  A ) system  S un iverse  T Since for spontaneous the entropy change of the universe must be positive, then (A) for the system must be negative. (A)system  0 (2) (G) System and (S) Universe - (  G ) system Δ S u n iverse  T Since for a spontaneous process the entropy change of the universe must be positive, (Gsystem) must be negative. (  G ) system  0 226 (i) –(A/T) is the entropy change of the universe when the change takes place at constant volume, (ii) –(G/T) is the entropy change of the universe when the system undergoes a change at constant pressure. (i) Criteria for equilibria at constant T and V. A = E – TS At constant temperature and volume for a reversible process (dA)T,V = 0 dA= - SdT- PdV For finite changes we will have ( A)T , V 0 7 227  12/4/2024 In an irreversible process (spontaneous changes) (dA) T,V  0 For finite changes ( ΔA) T, V  0 (ii) Criteria for equilibria at constant T and P. For this case we consider the Gibbs free energy function G = H – TS For a reversible change dG = -SdT + VdP 228 At constant temperature and pressure dT = 0 = dP and (dG)T,P = 0 For finite changes (G)T,P  0 For irreversible (spontaneous) changes (dG)T,P < 0 (ΔG)T , P  0 For finite changes 8 229  12/4/2024 The thermodynamic criteria of equilibrium are summarized below: Property Sign Nature of Process (G)T,P or (A)T,V Negative Spontaneous Zero Equilibrium Positive Non-spontaneous 230 Physical Equilibria Involving Phase Transitions: The Clapeyron-clausius equation: Clapeyron equation A thermodynamics relation between the changes of pressure with change of temperature of a system at equilibrium is called Clapeyron equation. This relationship can be derived as follows: 9 231  12/4/2024 Let us consider the equilibrium transition of one mole of a substance from one phase () to another phase () occurring at constant pressure and temperature. Phase () = phase (), at constant T and P. At equilibrium the free energy of the substance in both the phases must be identical, since (G)T,p = 0. That is G G 232 Let the temperature and pressure are changed to the new values (T + dT) and (P + dP) in such a manner that the phases  and  are still at equilibrium. Now, (G + dG) and (G + dG), then at equilibrium (G  dG) (G  dG) α β dG  SdT VdP  (dG)  (dG) α β  SdT  V dP   Sβ dT  Vβ dP 10 233  12/4/2024 (Sβ  Sα ) dT  (Vβ Vα ) dP dP Sβ  S ΔS Clapeyron   equation dT Vβ  Vα ΔV S = H/T, where H is the change in enthalpy for the reversible transformation at temperature T. Substituting this values of  S into Clapeyron equation we get, dP H Clapeyron dP  HT V dT T  V equation dT 234 dP H  dT T  V Above equation is integrated on the assumption that  H and  V are constant, P2 T2 H dT  dP   V  T P1 T1 H T2 (P2  P1 )  ln V T1 This form of equation is usually employed for processes involving condensed phases.(Solid change to liquid without present vapour phase) 11 235  12/4/2024 dP H Applications of Clapeyron Equation  Clapeyron dT T  V equation Liquid-vapour equilibria (Vaporization process) Liquid = Vapour Phase () = Phase ()  H =  Hvap = Enthalpy of vaporization  V = V - V = (Vv - VL) dP ΔHvap  dT T(VV  VL ) 236 Since  Hvap is positive and V > V dP dT will always be positive i.e., with increase of P, there must be increase in T. 12 237  12/4/2024 (ii) Solid-liquid equilibria (Fusion process). Solid = Liquid Phase () = Phase () H = Hfus = Enthalpy of fusion  V = Volume of liquid (VL) – Volume of solid (Vs) Clapeyron equation can be written as dP ΔH fus  dT T(V L  VS ) 238 For the conversion of solid into liquid heat is required and  Hfus is always positive. But the volume of liquid may be greater or less than that of solid. So the sign of dP/d T will depend on the nature of the system. For example, the volume of liquid water is less than the volume of ice, so V is negative and hence dP/dT, is negative. 13 239  12/4/2024 That is, with increase of pressure, the melting point would decrease for ice-water equilibrium. The same is true for bismuth metal. 240 (iii) Solid-vapour equilibrium (sublimation process) Solid = Vapour Phase () = Phase () H = Hsub = Enthalpy of sublimation.  Hsub  Hfus  Hvap V = volume of vapour phase – volume of solid phase. dP  H sub Hsub is always positive  (64) dT T  V V, the solid-vapour equilibria have always a positive value 14 241  12/4/2024 (iv) Equilibrium between two crystalline forms S o li d ( )  S o l i d (  ) dP Δ Htrans  (65) dT T (V  V ) where  Htrans = Enthalpy of transition, V and V are the molar volumes of the indicated form. The sign of dP/dT will depend upon the signs of  Htrans and the densities of the two forms. 242 The Clausius- Clapeyron Equation Clausius-Clapeyron equation is applied in case liquid vapour equilibria dP ΔH vap  The Clapeyron equation, dT T(V V  V L ) Vv >> VL and Vv = RT/P. dP Δ H vap  dT T V V dP ΔHvap P ΔHvap (66)   dT TVV RT2 15 243  12/4/2024 This differential equation may be employed if the changes in temperature and pressure are small. ΔT RT 2  (67) ΔP ΔH vap P 244 Integration of Clausius-Clapeyron Equation dP ΔHvap dT  (68) P R T2 Integration on the assumption on that  Hvap is independent of temperature, we get, ΔHvap ln P  T 2 dT Const. H vap R ln P    C (69) RT 16 245  12/4/2024 -. C being the integration constant. a plot of ln P versus 1/T should be linear with a slope of Hvap / R Thus, the heat of vaporization can be calculated,  H vap using (in Joules per mole) = Slope  8.314. 246 It is more convenient to integrate equation between suitable initial and final limits to get, P2 ΔHvap T2 2 Hvap 1 1  ln P  R  T dT lnP2 lnP1  [  ] R T2 T1 P1 T1 or P2 ΔHvap  T2  T1  ln ( )   P1 R  T1T2  This equation can be used to calculate the heat of vaporization if the vapour pressure data at two different temperatures are known. 17 247  11/28/2024 dq rev  T 0 2 dq dq rev rev S  S2  S1 dS  1 T T mathematical statements of the second law of thermodynamics. T2 V V  Isothermal S=nCVln  nRln 2 ΔS  nRln  2  T1 V1  V1  S=nCPln T2 P  nRln 1 P  T1 P2 ΔS  nR ln  1   T2  reversible adiabatic Reversible Adiabatic Changes  S  0 processes are called ΔH trans isentropic processes. Phase changes. ΔS trans  Ttrans Entropy Change for Isolated System (dS)rev = 0 and (dS)irr > 0 Or (S)rev = 0 and (S)irr > 0 1  11/28/2024 Entropy Change on Heating or Cooling of a Substance T  T  ΔS  nC V ln  2  ΔS  nC P ln  2   T1   T1  ΔSuniverse  ΔSsystem  ΔSsurroundings Example 1 mole of an ideal gas is compressed isothermally at 298oK. If the gas is compressed from 1 atm to 10 atm reversibly and irreversibly against an external pressure of 10 atm, calculate the entropy change for both cases. ΔS For the reversible isothermal process P 1 S system  nR ln  1   1*8.314 ln    19.14 EU  P2   10  ΔS (surroundings) = +19.14 EU ΔS (universe) = ΔS (system) + ΔS (surroundings) = -19.14 + 19.14 = 0 2  11/28/2024 ΔS For the irreversible isothermal process ΔS (system) in the irreversible = reversible ΔS (system) Because the Entropy is a state function ΔSsystem,irreversible= -19.14 EU  P   10  q  nRT 1  2   1* 8.314 * 2981    1* 8.314 * 298 * 9 irreversib le  P   1  1 q  22298.15 J irreversib le This is the heat lost by the system and gained by the surrounding qsurroundings,irreversible = + 22298.15 J ΔSsurroundings,irreversible = qsurr/T= +22298.15/298= +74.9 EU ΔS (universe) = ΔS (system) + ΔS (surroundings) =-19.14 +74.9 = +55.83 EU Example What is the change in entropy when argon at 25oC and 1 atm pressure is expanded isothermally from 500 cm3 to 1000 cm3 and simultaneously heated to a temperature of 100oC, CV=3/2 * R PV 1* 0.5 n   0.020 moles RT 0.082 * 298 V  T  S  nR ln  2   nCV ln  2   V1   T1   1000  3  373  S  0.020 *8.314* ln    0.020* *8.314 ln    500  2  298  S  0.1153 .05599  0.172 J / K 3  11/28/2024 Combined Form of the First and Second Laws of thermo-dynamics The mathematical form of the first law of thermodynamic is, dq = dE + dw = dE + PdV The mathematical expression for second law is dq rev or dqrev = TdS dS  T substitution of equation in the first law equation give, TdS = dE + PdV dE = TdS – PdV combined form of the first and the second laws of thermodynamics. According to equation dE = T dS - PdV  E   E    T (1)    P (2)  S  V  V S Differentiate (1) with respect to V at differentiate equation (2) with respect to S at constant S constant V 2E  T     2E  P VS  V S    SV  SV  2E 2E  T   P         Maxwell equation(1) VS SV  VS  SV 4  11/28/2024 Enthalpy and Entropy: Enthalpy is expressed as H = E + PV dH = dE + PdV + VdP using equation , dE = TdS – PdV dH = TdS – PdV + PdV + VdP  dH = TdS + VdP dH = TdS + VdP H H    T (1)    V (2)  S P  PS Differentiate (1) with respect to P differentiate equation (2) with respect to S at at constant S constant P 2H  T   2H  V        PS  P S SP  S  P  T V Maxwell equation (2)     PS  SP 5  11/28/2024 Energy as a Function of S and V dE = T dS - PdV E = f (S,V) The total differential of E is given as  E   E  dE    dS    dV  S  V  V S Comparing the coefficients with this equation and dE = T dS - PdV E  E    T    P  SV  V S  E  Equation   T can be written as  S  dE = TdS For a given temperature this equation can be integrated to give a new expression + Integration constant  dE  T  dS E = TS + A Or A = E – TS A = Integration constant and called Helmholtz free energy function. 6  11/28/2024 Enthalpy as a Function of S and P From equation dH = TdS + VdP H = f(S, P)  H   H  H  H dH    dS    dP  S  p  P  S   T   V  Sp  P S  dH   TdS Constant (G) H = TS + G The function G is the same as the G = H – TS Gibbs free energy function Chapter V 7  11/28/2024 We have studied three thermodynamic function for any system; these are the energy content (E), the heat content (H) and the entropy (S). Although the entropy concept is the fundamental consequence of the second law of thermodynamics, there are two other function, which utilize the entropy in their derivation. The free energy (G) will be employed extensively in connection with the study of equilibria, both physical and chemical, and the direction of chemical change. The second is called the work function (A)or Helmholtz free energy function. 8  11/28/2024 Free Energy Functions A and G Entropy is a measure of unavailable energy. entropy is multiplied by absolute temperature the product (TS) is equal to the amount of heat (energy) which is not free to be used for useful work. Then out of total heat absorbed by a system the amount less by TS must be the available amount of heat which can be put for useful work. Thus the free energy (X) can be expressed as (X) = q – TS (1) Since heat is absorbed either at constant V or at constant P the term q may be replaced either by E or H. Therefore, the free energy would be equal to E – TS or H-TS. Two new symbols are given to these free energy terms. 9  11/28/2024 1-Helmholtz free energy A A = E - TS 2- Gibbs free energy G G = H - TS Properties of A 1- A = E – TS dA = dE – TdS – SdT (Differential change) But dE = dq – dw dE = TdS – dw dA = TdS – dw – TdS – SdT = -dw – SdT If T is constant dT = 0, and –(dA)T = dw The decrease in Helmboltz free energy is or -(A)T = wrev equal to the maximum reversible work which a system can be do isothermally. 10  11/28/2024 2- A = E – TS dA = dE – TdS – SdT = TdS-dw – TdS -SdT dA = -PdV – SdT At constant volume, dV = 0 At constant temperature, dT = 0  A   A     S   P  T V  V T Differentiating with respect to V at constant T Differentiating with respect to T at constant V 2 2A  P  A  S       TV  TV V T  V T 2A 2A P  S applying Euler's theorem      Maxwell`s equation (3) TV VT Tv VT P  S     Maxwell`s equation (3) Tv VT This equation is very useful to study the variation of entropy with volume at constant temperature. Experimentally it is impossible to measure the change of entropy with change of volume as we do not have any apparatus which can measure entropy directly. But it is convenient to study the variation of pressure with temperature at constant volume. Hence, using Maxwell's equation, the experimentally difficult quantities can be calculated easily 11  11/28/2024 3- A = E – TS dA = TdS – dw – TdS – SdT (dA)T = - dW (dA)T = -PdV For ideal gas, nRT P V dV dA nRT V V  P A  nRT ln  2   nRT ln 1 Integration gives  V1  P2 4 - Variation of A with T at constant V A = E – TS   A  E     2 T  T  v T 12  11/28/2024 4 - Variation of A with T at constant V A = E – TS dA = dE – TdS – SdT dA = -SdT – PdV A but  S  T  A   A  E  T  (13)  T v  A  or A  T   E  T  v 1 Multiplying both sides by  T2 we get  A   A  T   T  v E   T2 T2 13  11/28/2024  A 1 A E  ( ) V  (14) T2 T T T2   A L.H.S. is the differential of equation 14   and hence T  T   A E     (15) T  T  v T2 Consider a system in state 1. Let it is changed isothermally to state 2. Thus equation 13 changed to, A  A1 E1 T 1   T V  A  and A2  E2  T  2   T V The change in A is given as 14  11/28/2024  A2 A1 A(E2 E1)T  A2 A1V T   A    E  T     T V   A  Or A  T    E  T  V 1 On multiplying by  and replacing L.H.S. by differential T2   A    form, we get T  T    A  E (16)     T  T  v T2 Properties of G 1.G = H – TS dG = dH –SdT – TdS H = E + PV dH = dE + PdV + VdP  dG = dE + PdV + VdP – SdT - TdS but dE = dq – dw = TdS - dwrev  dG = TdS – dw + PdV + VdP – SdT – TdS  dG = – dw + PdV + VdP – SdT At constant temperature and pressure dT = 0 = dp, (dG)T,P = -(dw-PdV) 15  11/28/2024 (dG)T,P = -(dw-PdV) where dw includes all types of work. PdV = mechanical work and therefore, dw-PdV = dw' = net work = useful work Thus, -(dG)T,P = dwnet - (G)T,P = wnet The decrease in free energy is equal to the reversible maximum work which a system can be isothermally and isobarically over and above the mechanical work. 2. G = H-TS dG = dH-TdS-SdT But H=E+PV dH=dE+PdV+VdP dG = TdS – dw + PdV + VdP – SdT – TdS dG = TdS – PdV + PdV + VdP – SdT – TdS  dG = -SdT+VdP 16  11/28/2024 dG = -SdT+VdP At constant P At constant T  G  G    S   V  P T  T P Applying Euler's theorem for dG  2G  S   2G   V              PT T  P T  TP P  T P  S   V       Maxwell's (4) P  T  T P 1.Variation of free energy with pressure at constant T: dG = -SdT+VdP At constant temperature, dT = 0, the equation will reduced to (dG)T =VdP If the state of the system is changed from 1 to 2, then 2 P2 nRT P2 dP  dG   VdP V G2 G1 G  nRT  1 P1 P P1 P P2 V1 G  nRT ln G  nRTln P1 V2 This equation can be used to calculate the free energy for expansion or compression. 17  11/28/2024 4. Calculation of Free Energy Changes with temperature at constant Pressure Derivation of Gibbs-Helmholtz Equation Gibbs-Helmholtz equation relates the free energy change to the enthalpy change and the rate of change of free energy with temperature. G=H-TS   (G / T)   H  T   P T2 There are two forms of Gibbs-Helmholtz equation. These are (1) The relation between E and A is given by equations A = E -TS   A E    T  T v T2 (2) The relation between G and H, will be derived here. Consider the relation, G=H-TS ,   ( G / T)   H    2  T P T 18  11/10/2024 Carnot Heat Engine Total work w cyc q 2  q1    Heat absorbed q 2 q2 2 T2  T1 T q   1 1 1 1 T2 T2 q2 0 < < 1 Carnot’s theorem: No engine operating between 1two heat reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. Entropy q1 T q1 T   1 1 1    1 q2 T2 q2 T2 q1 T  1  0 q2 T2 Multiplying by q2 / T1 we obtain, q1 q it suggests that is a state  2  0 T1 T2 function S define by 2  11/10/2024 d q rev ds  T where the subscript is required because we are considering a reversible carnot engine. The change in state function around any cycle is zero and this is true for S this new state function is called entropy. dq rev  T 0  Scy =  S1 +  S2 +  S3 +  S4 q2 q1  0   00 T2 T1 There is no change in entropy in the reversible d q rev ds  adiabatic steps. T Then can be written as follows for a general cyclic process. dq rev  T 0 3  11/10/2024 The Concept of Entropy: According to the second statement of the law, heat cannot be completely converted into work. Why? Where does the remaining amount of heat go which cannot be converted into work? What is that factor which is a measure of the extra change other than the work done? Clausius invented a term S, which is measure of this extra work. This term is called entropy (trope = change, a suffix ‘en’ is written to identify it with energy). Process (a) w1 Consider a gas present in a cylinder fitted with an ideal piston. If q is the amount of heat supplied to the gas, heat (a) the gas will expand and the piston will be pushed. The work done by the gas = P  V w2 Process (b). If the work equivalent to the above amount of work is done on the gas, the gas would be compressed and heat heat (b) will be evolved. 4  11/10/2024 Experimentally it has been observed that the amount of heat evolved in the process B is less than the amount of heat absorbed in process A. This clearly indicates that the whole of the heat absorbed in process A has not been converted into work. Then the question arises, where has the heat been utilized? When heat is absorbed by the gas, the kinetic energy of the gas molecules will be increased and they will move faster and collide with (I) each other, (ii) the container and (iii) the face of the piston, thereby the piston will pushed. Only the third type of collision will bring about the expansion of the gas The first and second type of collisions will not bring effective expansion of the gas. Thus some collisions and hence a fraction of heat is not utilized for the expansion of the gas. Certainly the work due to expansion is not exactly equivalent to the amount of heat absorbed. Since the heat goes waste only due to the random motion of the gas molecules, so entropy is a measure of randomness 5  11/10/2024 Presentation of Entropy Concept Mathematical Treatment: Let us consider an ideal gas in reversible change from initial state T1, V1 to a final state T2, V2 the first law of thermodynamics can be written as. dqrev= dE + dwrev nRT dwrev = PdV PV = nRT P V dqrev = dE + PdV For an ideal gas, dE = nCV dT dV d q re v = n C V d T + n R T V Dividing throughout by T, we get dq rev dT dV 2 T V2  nCV  nR dq rev 2 dT dV T T V 1 T  nC V   nR  T1 T V1 V 2 dqrev T V   nCV ln 2  nRln 2 1 T T1 V1 P1V1 P2V 2 V 2 P1T 2   T1 T2 V1 P2 T1 Substitute in the above equation 6  11/10/2024 2 dq rev T T P   nC V ln 2  nR ln 2  nR ln 1 1 T T1 T1 P2 2 dq rev T P 2 dqrev T P   n(CV +R) ln 2  nR ln 1  1  nCP ln 2  nR ln 1 1 T T1 P2 T T1 P2 2 dq dq rev rev dS  S  S2  S1 T 1 T The last two equation are mathematical statements of the second law of thermodynamics. Thus, dq is a path function it becomes a state function when multiplied by an integrating factor 1 T Reversible Isothermal Volume Change of Ideal Gases T2 V S=nCVln  nRln 2 T1 V1 Since for the isothermal expansion T constant, the entropy change of the system can be evaluated, V2 S= nRln V1 7  11/10/2024 Since at constant temperature P1V1 = P2V2 then V2/V1 = P1/P2, isothermal expansion from pressure P1 to P2 P  P S  nR ln  2   nR ln 1  P1  P2 Entropy Change on Heating or Cooling of a Substance Let us consider n moles of a substance at a temperature T1. Let its temperature is changed to T2 keeping the volume constant. If this change is brought about reversibly T2 V T  S=nCVln  nRln 2 ΔS  nC V ln  2  T1 V1  T1  8  11/10/2024 For temperature changes for T1 to T2 at constant P, T2 P T2 ΔS  nCP ln  nRln 1 ΔS  nCP ln T1 P2 T1 Reversible Adiabatic Changes If any system undergoes a reversible adiabatic change, then by definition, dqrev = 0 at all stages of the process, so that S  0 Therefore, reversible adiabatic processes are often called isentropic processes. Phase changes. As an example of phase changes the melting of a solid. Solid Liquid This process can be considered as reversible and entropy change can be given as S  H f Tf In general it is clear that for any phase transformation taking place at constant temperature and pressure, ΔH trans ΔS trans  Ttrans 9  11/10/2024 where Htrans is the enthalpy change of the transformation (e.g. melting, vaporization etc.) and Ttrans is the constant temperature at which the transformation takes place. Entropy Change for Isolated System For reversible isolated processes the change of entropy is zero where as for irreversible isolated process the entropy change is greater than zero. In an isolated system there is no exchange of energy, matter, heat etc., across the boundary. Thus dq = 0, dw = 0, for an isolated system. (dS)rev = 0 and (dS)irr > 0 Or (S)rev = 0 and (S)irr > 0 10  11/10/2024 the total entropy is called the entropy of the universe, defined as ΔSuniverse  ΔSsystem  ΔSsurroundings Example Five moles of an ideal gas is expanded from initial volume of 10 liters to a final volume of 20 liters isothermally at 300K calculate the entropy changes of : a) System. b) Surroundings. c) Universe V Δ S  n R ln 2 V 1  5  8.3  ln 20 28.81JK1 10 ΔS  28.81JK1 sys ΔS   ΔS ΔS ΔS ΔS  0 surr sys universe sys surr   28.81 11  11/10/2024 Example Homework An ideal gas at initial pressure of 20 atm and 300oK occupies a volume of 1 liter calculate the entropy of the system for reversible isothermal expansion of the gas to a final volume 10 liters. Example Evaluate the change in entropy when 3 moles of a gas are heated from 27oC to 727oC at a constant pressure of 1 atm. The heat capacity of the gas is 23.7 JK-1. T2  S  nCp ln T1 1000  3 x 23.7 ln 300  85.5 JK 1 12  11/10/