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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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