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Thermodynamics

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  1. http:\\asadipour.kmu.ac.ir...43 slides

  2. Thermodynamics http:\\asadipour.kmu.ac.ir...43 slides

  3. Thermodynamics 1- Spontaneous Processes 2- The Definition of Entropy 3- Entropy and Physical Changes 4- Entropy and the Second Law of Thermodynamics 5- The Effect of Temperature on Spontaneity 6- Free Energy 7- Entropy Changes in Chemical Reactions 8- Free Energy and Chemical Reactions 9- Free Energy and Equilibrium http:\\asadipour.kmu.ac.ir...43 slides

  4. Kinetic: predicts speed (rate). Thermodynamic: predicts direction . Spontaneous Processes without outside intervention maybe slowly Cdiamond→Cgrafit http:\\asadipour.kmu.ac.ir...43 slides

  5. First Law of Thermodynamics The change in the internal energy of a system is equal to the work done on it plus the heat transferred to it. The Law of Conservation of Energy E = q - w Second Law of Thermodynamics For a spontaneous process the Entropy of the universe (meaning the system plus its surroundings) increases. Suniverse> 0 Third Law of Thermodynamics In any thermodynamic process involving only pure phases at equilibrium andcrystalline substance S, approaches zero at absolute zero temperature; S = 0 at 0 K http:\\asadipour.kmu.ac.ir...43 slides

  6. First Law of Thermodynamics • Energy cannot be created nor destroyed. • Therefore, the total energy of the universe is a constant. • Energy can be converted from one form to another • or transferred from a system to the surroundings or vice versa. http:\\asadipour.kmu.ac.ir...43 slides

  7. First Law of Thermodynamics • E of adiabatic system is constant & incomputable. • E=Ef-Ei • Ef=Ei+q-W • Ef-Ei=q-W • E= q-W • E= state function http:\\asadipour.kmu.ac.ir...43 slides

  8. First Law of Thermodynamics • E= q-W W=PV • E= q -PV • qp=E+PV • H= E+PV • qp=H • H =E+PV • PV=nRT • H =E+nRT http:\\asadipour.kmu.ac.ir...43 slides

  9. First Law of Thermodynamics • R=Gases constant= 0.082 L.atm/mol.Kº • R=Gases constant= 8.314 J/mol.Kº http:\\asadipour.kmu.ac.ir...43 slides

  10. What is a spontaneous process? not probable probable This is not a spontaneous process. The reverse process (going from right to left) is spontaneous. http:\\asadipour.kmu.ac.ir...43 slides

  11. Summary of Entropy • Entropy is the degree of randomness or disorder in a system Ssolid < S liquid < Sgas • The Entropy of all substances is positive • ΔSsys =ΔSsis the Entropy Change of the system • ΔSsur = ΔSeis the Entropy Change of the surroundings • ΔSuni =ΔStis the Entropy Change of the universe • S has the units J K-1mol-1 http:\\asadipour.kmu.ac.ir...43 slides

  12. For gas, liquid or solid, If T S When a liquid vaporizes, S When a liquid freezes, S S and T • T<0 H2O (l)→ H2O (s) S<0 Spontaneous reaction http:\\asadipour.kmu.ac.ir...43 slides

  13. T<0 H2O (l)→ H2O (s) S<0 Suniverse=Ssystem + Ssurroundings (St=Ss +Se) ΔSuniv > 0 Spontaneous Forward ΔSuniv = 0 At Equilibrium ΔSuniv< 0 Spontaneous Reverse http:\\asadipour.kmu.ac.ir...43 slides

  14. http:\\asadipour.kmu.ac.ir...43 slides

  15. Suniverse=Ssystem + Ssurroundings • The magnitude of ΔSsur depends on the temperature This is ΔH of the system. If the reaction is exothermic, ΔH has a negative sign and ΔSsurr is positive If the reaction is endothermic, ΔH has a positive sign and ΔSsurr is negative http:\\asadipour.kmu.ac.ir...43 slides

  16. Ssystem + Ssurroundings =Suniverse http:\\asadipour.kmu.ac.ir...43 slides

  17. Entropy • Like total energy, E, and enthalpy, H, entropy is a state function. • Therefore, S = SfinalSinitial http:\\asadipour.kmu.ac.ir...43 slides

  18. Gibbs Free Energy • St =Ss +Se • St =Ss - H/T • TSt=TSs - H • -TSt=-TSs+ H • -TSt= H -TSs • G=H-TS • G= H -TSs http:\\asadipour.kmu.ac.ir...43 slides

  19. Free Energy and Chemical Reactions ΔG = ΔH - T·ΔS W q ΔG TΔS ΔH Operating cell Ideal reverse cell Spontaneous reaction http:\\asadipour.kmu.ac.ir...43 slides

  20. Entropy ΔSuniv > 0 Spontaneous Forward ΔSuniv = 0 Equilibrium ΔSuniv < 0 Spontaneous Reverse Free Energy ΔG < 0 Spontaneous ΔG = 0 Equilibrium ΔG > 0 Spontaneous Reverse http:\\asadipour.kmu.ac.ir...43 slides

  21. Effects of Temperature on ΔG° ΔG° = ΔH° - TΔS° • Typically ΔH and ΔS are almost constant over a broad range • For the reaction above, as Temperature increases ΔG becomes more positive, i.e., less negative. 3NO (g) → N2O (g) + NO2 (g) http:\\asadipour.kmu.ac.ir...43 slides

  22. Spontaneity of reactions in different temperatures ΔG = ΔH - T·ΔS Case D ΔH° < 0 ΔS° < 0 ΔG = ΔH - T·ΔS ΔG < 0 or spontaneous at low Temp. Case C ΔH° > 0 ΔS° < 0 ΔG = ΔH - T·ΔSΔG > 0 or non-spontaneous at all Temp. B A Case A ΔH° > 0 ΔS° > 0 ΔG = ΔH - T·ΔSΔG < 0 or spontaneous at high Temp. Case B ΔH° < 0 ΔS° > 0 ΔG = ΔH - T·ΔS ΔG < 0 spontaneous at all temp. C D http:\\asadipour.kmu.ac.ir...43 slides

  23. Entropies of Reaction ΔSrxn° = ΣS°products – ΣS°reactants ΔSrxn° is the sum of products minus the sum of the reactants, for one mole of reaction (that is what ° means) For a general reaction a A + b B → c C + d D values, S° in units JK-1mol-1 http:\\asadipour.kmu.ac.ir...43 slides

  24. Standard Entropies • Standard entropies • tend to increase with increasing molar mass. • MW S http:\\asadipour.kmu.ac.ir...43 slides

  25. http:\\asadipour.kmu.ac.ir...43 slides

  26. Calculate ΔSr° at 298.15 K for the reaction 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) S°SO2(g) =(248) JK-1mol-1 S° H2O(g)= (189) JK-1mol-1 S°H2S(g) = (206) JK-1mol-1 S° O2(g) = (205) JK-1mol-1 Solution ΔSrxn°= 2S°SO2(g) + 2S°H2O(g) -2S°H2S(g) - 3S°O2(g) ΔSrxn°= 2(248) + 2(189) -2(206) - 3(205) ΔSrxn°= -153 JK-1mol-1 mol= mol of reaction(total reaction=1 mol) http:\\asadipour.kmu.ac.ir...43 slides

  27. (b) Calculate ΔS° when 26.7 g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at 298.15K ΔSrxn°= -153 JK-1mol-1 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) Solution: http:\\asadipour.kmu.ac.ir...43 slides

  28. Free Energy and Chemical Reactions ΔG = ΔH - T·ΔS • Because G is a State unction, for a general reaction a A + b B → c C + d D http:\\asadipour.kmu.ac.ir...43 slides

  29. Calculate ΔG° for the following reaction at 298.15K. Use texts for additional information needed. 3NO(g) → N2O(g) + NO2(g) • ΔGf°(N2O) = 104 kJ mol-1 • ΔGf°(NO2) = 52 • ΔGf° (NO) = 87 ΔG°= 1(104) + 1(52) – 3(87) ΔG°= − 105 kJ therefore, spontaneous http:\\asadipour.kmu.ac.ir...43 slides

  30. The Dependence of Free Energy & Pressure ΔG = ΔG° + RT ln Q Where Q is the reaction quotient • a A + b B⇋c C + d D • If Q = KΔG = 0((equilibrium)) ΔG° = -RT ln K • If Q > KΔG = + the rxn shifts towards the reactant side • If Q < KΔG = - the rxn shifts toward the product side compare http:\\asadipour.kmu.ac.ir...43 slides

  31. a A + b B⇋c C + d D • At Equilibrium conditions, ΔG = 0 • ΔG° = -RT lnK • NOTE: • we can now calculate equilibrium constants (K) for reactions from standard ΔGf functions of formation http:\\asadipour.kmu.ac.ir...43 slides

  32. Spontaneous reactions • ΔG = ΔG° + RT ln Q • Where Q is the reaction quotient • a A + b B ↔ c C + d D http:\\asadipour.kmu.ac.ir...43 slides

  33. ΔGrxn°= – 105 kJ mol-1 Calculate the equilibrium constant for this reaction at 25C. 3NO(g) ↔ N2O(g) + NO2(g) • Solution UseΔG ° =- RT ln K Rearrange Compare http:\\asadipour.kmu.ac.ir...43 slides

  34. ΔG° and KeqΔG = ΔG° + RT ln QIn equilibrium ΔG =0 ΔG° = - RT lnKeq http:\\asadipour.kmu.ac.ir...43 slides

  35. Gibbs Free Energy & equilibrium N2+3H2 ⇋3NH3ΔG°=+5KJ • If DG is negative, the forward reaction is spontaneous. • If DG is 0, the system is at equilibrium. • If G is positive, the reaction is spontaneous in the reverse direction. Is the reaction spontaneoues at 25 0C http:\\asadipour.kmu.ac.ir...43 slides

  36. The Temperature Dependence of Equilibrium Constants • ΔG = ΔH - T·ΔS • Divide by RT, then multiply by -1 http:\\asadipour.kmu.ac.ir...43 slides

  37. EXothermic R. T K • Notice that this is y = mx +b • the equation for a straight line KT • A plot of y = mx + b or • ln K vs. 1/T Endothermic R. T K K  T http:\\asadipour.kmu.ac.ir...43 slides

  38. If we have two different Temperatures and K’s ΔH and ΔS are constant over the temperature range http:\\asadipour.kmu.ac.ir...43 slides

  39. The reaction 2 Al3Cl9 (g) → 3 Al2Cl6 (g) Has an equilibrium constant of 8.8X103 at 443K and a ΔHr°= 39.8 kJmol-1 at 443K. Estimate the equilibrium constant at a temperature of 600K. http:\\asadipour.kmu.ac.ir...43 slides

  40. http:\\asadipour.kmu.ac.ir...43 slides

  41. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. http:\\asadipour.kmu.ac.ir...43 slides

  42. Standard Entropies • Standard entropies • tend to increase with increasing molar mass. • MW S http:\\asadipour.kmu.ac.ir...43 slides

  43. Benzene, C6H6, (at 1 atm) ,boils at 80.1°C and ΔHvap = 30.8 kJ • A) Calculate ΔSvap for 1 mole of benzene at 60°C and pressure = 1 atm. • ΔGvap=ΔHvap-TΔSvap • at the boiling point, ΔGvap= 0ΔHvap= TbΔSvap • B) Does benzene spontaneously boil at 60°C? http:\\asadipour.kmu.ac.ir...43 slides