1. 1911209http:\\asadipour.kmu.ac.ir...46 slides

2. Thermodynamics 2911209http:\\asadipour.kmu.ac.ir...46 slides

3. Thermodynamics 1 Spontaneous Processes 2 The Isothermal Expansion and Compression of an Ideal Gas 3 The Definition of Entropy 4 Entropy and Physical Changes 5 Entropy and the Second Law of Thermodynamics 6 The Effect of Temperature on Spontaneity 7 Free Energy 8 Entropy Changes in Chemical Reactions 9 Free Energy and Chemical Reactions 10 The Dependence of Free Energy on Pressure 11 Free Energy and Equilibrium 3911209http:\\asadipour.kmu.ac.ir...46 slides

4. –Thermodynamic: – predicts direction. –Kinetic: –predicts speed (rate). Spontaneous Processes –without outside intervention –maybe slowly –C diamond → C grafit 4911209http:\\asadipour.kmu.ac.ir...46 slides

5. 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.  S universe > 0 Third Law of Thermodynamics In any thermodynamic process involving only pure phases at equilibrium and crystalline substance  S, approaches zero at absolute zero temperature;  S = 0 at 0 K 911209http:\\asadipour.kmu.ac.ir...46 slides

6. First Law of Thermodynamics Energy cannot be created nor destroyed. Therefore, the total energy of the universe is a constant. Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. 6911209http:\\asadipour.kmu.ac.ir...46 slides

7. First Law of Thermodynamics E of adiabatic system is constant & incomputable.  E=E b -E a E f =E i +q-W E f -E i =q-W  E= q-W  E= state function 7911209http:\\asadipour.kmu.ac.ir...46 slides

8. First Law of Thermodynamics W=P  V  E= q -P  V q p =  E+P  V H= E+PV q p =  H  H =  E+P  V PV=nRT  H =  E+  nRT 8911209http:\\asadipour.kmu.ac.ir...46 slides

9. First Law of Thermodynamics R=Gases constant= 0.082 L.atm/mol.Kº R=Gases constant= 8.314 J/mol.Kº 9911209http:\\asadipour.kmu.ac.ir...46 slides

10. probable not probable This is not a spontaneous process. The reverse process (going from right to left) is spontaneous. What is a spontaneous process? 10911209http:\\asadipour.kmu.ac.ir...46 slides

11. Summary of Entropy Entropy is the degree of randomness or disorder in a system The Entropy of all substances is positive S solid < S liquid < S gas ΔS sys = ΔS s is the Entropy Change of the system ΔS sur = ΔS e is the Entropy Change of the surroundings ΔS uni = ΔS t is the Entropy Change of the universe S has the units J K -1 mol -1 11911209http:\\asadipour.kmu.ac.ir...46 slides

12. For gas, liquid or solid, If T  S  When a liquid vaporizes, S  When a liquid freezes, S  12 T<0 H 2 O (l) → H 2 O (s)  S<0 Spontaneous reaction S and T 911209http:\\asadipour.kmu.ac.ir...46 slides

13. 13 –T<0 H 2 O (l) → H 2 O (s)  S<0  S universe =  S system +  S surroundings (  S t =  S s +  S e ) ΔS univ > 0Spontaneous Forward ΔS univ = 0At Equilibrium ΔS univ < 0Spontaneous Reverse StSt SeSe SsSs T -0.08+22.05-22.13+1 0+21.99-21.990 +0.08+21.93-21.85 911209http:\\asadipour.kmu.ac.ir...46 slides

14. 14 1 st Law of Thermodynamics –In any process, – the total energy of the universe remains unchanged: – energy is conserved –A process and its reverse are equally allowed by the first law ΔE forward + ΔE reverse =0 (Energy is conserved in both directions) 2 nd Law of Thermodynamics –Processes that increase ΔS universe are spontaneous. 911209http:\\asadipour.kmu.ac.ir...46 slides

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16. The magnitude of ΔS sur depends on the temperature If the reaction is exothermic, ΔH has a negative sign and ΔS surr is positive If the reaction is endothermic, ΔH has a positive sign and ΔS surr is negative  S universe =  S system +  S surroundings This is ΔH of the system. 16911209http:\\asadipour.kmu.ac.ir...46 slides

17.  S system +  S surroundings =  S universe 17911209http:\\asadipour.kmu.ac.ir...46 slides

18. Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial 18911209http:\\asadipour.kmu.ac.ir...46 slides

19. Gibbs Free Energy  S t =  S s +  S e  S t =  S s -  H/T T  S t =T  S s -  H -T  S t =-T  S s +  H -T  S t =  H -T  S s G=H-TS  G=  H -T  S s 19911209http:\\asadipour.kmu.ac.ir...46 slides

20. Free Energy ΔG < 0Spontaneous ΔG = 0Equilibrium ΔG > 0Spontaneous Reverse Entropy ΔS univ > 0Spontaneous Forward ΔS univ = 0Equilibrium ΔS univ < 0Spontaneous Reverse 20911209http:\\asadipour.kmu.ac.ir...46 slides

21. Effects of Temperature on ΔG° 3NO (g) → N 2 O (g) + NO 2 (g) 21911209http:\\asadipour.kmu.ac.ir...46 slides

22. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS AB CD 22911209http:\\asadipour.kmu.ac.ir...46 slides

23. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS AB C D Case B ΔH° < 0 ΔS° > 0 ΔG = ΔH - T·ΔS ΔG = ( - ) - T·(+) = negative  ΔG < 0 or spontaneous at all temp. Case C ΔH° > 0 ΔS° < 0 ΔG = ΔH - T·ΔS ΔG = (+) - T·( - ) = positive  ΔG > 0 or non-spontaneous at all Temp. 23911209http:\\asadipour.kmu.ac.ir...46 slides

24. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS AB C D Case D ΔH° < 0 ΔS° < 0 ΔG = ΔH - T·ΔS at a low Temp ΔG = (-) - T·(-) = negative  ΔG < 0 or spontaneous at low Temp. Case A ΔH° > 0 ΔS° > 0 ΔG = ΔH - T·ΔS ΔG = ( + ) - T·( + ) at a High Temp  ΔG < 0 or spontaneous at high Temp. 24911209http:\\asadipour.kmu.ac.ir...46 slides

25. Entropies of Reaction ΔS rxn ° = ΣS° products – ΣS° reactants ΔS rxn ° 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 -1 mol -1 25911209http:\\asadipour.kmu.ac.ir...46 slides

26. Standard Entropies Standard entropies tend to increase with increasing molar mass. MW  S  26911209http:\\asadipour.kmu.ac.ir...46 slides

27. 27911209http:\\asadipour.kmu.ac.ir...46 slides

28. (a)Calculate ΔS r ° at 298.15 K for the reaction 2H 2 S(g) + 3O 2 (g) → 2SO 2 (g) + 2H 2 O(g) S°SO 2 (g) =(248) JK -1 mol -1 S H 2 O(g)= (189) JK -1 mol -1 S°H 2 S(g) = (206) JK -1 mol -1 S° O 2 (g) = (205) JK -1 mol -1 Solution ΔS rxn °= 2S°(SO 2 (g) ) + 2S°(H 2 O(g)) -2S°(H 2 S(g) ) - 3S°(O 2 (g)) ΔS rxn °= 2(248) + 2(189) -2(206) - 3(205) ΔS rxn = -153 JK -1 mol -1 = – 153 JK -1 mol -1 28911209http:\\asadipour.kmu.ac.ir...46 slides

29. (b) Calculate ΔS° when 26.7 g of H 2 S(g) reacts with excess O 2 (g) to give SO 2 (g) and H 2 O(g) and no other products at 298.15K 2H 2 S(g) + 3O 2 (g) → 2SO 2 (g) + 2H 2 O(g) Solution: 29911209http:\\asadipour.kmu.ac.ir...46 slides

30. Free Energy and Chemical Reactions ΔG = ΔH - T·ΔS Because G is a State unction, f or a general reaction a A + b B → c C + d D 30911209http:\\asadipour.kmu.ac.ir...46 slides

31. Free Energy and Chemical Reactions ΔG = ΔH - T·ΔS 31 W W q q ΔGΔG ΔGΔG ΔHΔH ΔHΔH TΔSTΔS TΔSTΔS Spontaneous reaction Ideal reverse cell Operating cell 911209http:\\asadipour.kmu.ac.ir...46 slides

32. 32 Calculate ΔG° for the following reaction at 298.15K. Use texts for additional information needed. 3NO(g) → N 2 O(g) + NO 2 (g) ΔG°= 1(104) + 1(52) – 3(87) ΔG°= − 105 kJ therefore, spontaneous ΔG f °(N 2 O) = 104 kJ mol -1 ΔG f °(NO 2 ) = 52 ΔG f ° (NO) = 87 911209http:\\asadipour.kmu.ac.ir...46 slides

33. 33 ΔG = ΔG° + RT ln Q Where Q is the reaction quotient a A + b B ↔ c C + d D If Q > K the rxn shifts towards the reactant side –The amount of products are too high relative to the amounts of reactants present, and the reaction shifts in reverse (to the left) to achieve equilibrium If Q = K equilibrium If Q < K the rxn shifts toward the product side –The amounts of reactants are too high relative to the amounts of products present, and the reaction proceeds in the forward direction (to the right) toward equilibrium compare The Dependence of Free Energy on Pressure 911209http:\\asadipour.kmu.ac.ir...46 slides

34. 34 At Equilibrium conditions, ΔG = 0 ΔG° = -RT ln K NOTE: we can now calculate equilibrium constants (K) for reactions from standard ΔG f functions of formation ΔG = ΔG° + RT ln Q –Where Q is the reaction quotient a A + b B ↔ c C + d D If Q < K the rxn shifts towards the product sideΔG ↓ If Q = K equilibrium ΔG =0 If Q > K the rxn shifts toward the reactant side ΔG ↑ 911209http:\\asadipour.kmu.ac.ir...46 slides

35. Calculate the equilibrium constant for this reaction at 25C. 3NO(g) ↔ N 2 O(g) + NO 2 (g) Solution Use ΔG ° =- RT ln K Rearrange 35911209 ΔG rxn °= – 105 kJ mol -1 http:\\asadipour.kmu.ac.ir...46 slides

36. 36 ΔG = ΔG° + RT ln Q Where Q is the reaction quotient a A + b B ↔ c C + d D Spontaneous reactions 911209http:\\asadipour.kmu.ac.ir...46 slides

37. 37 ΔG° and K eq ΔG = ΔG° + RT ln Q In equilibrium ΔG =0 ΔG° = - RT ln K eq  Gº(KJ) K -2001.1×10 35 -1003.3×10 17 -505.7×10 8 -252.4×10 4 -57.5 01.0 50.13 254.2×10 -5 501.7×10 -9 1003.0×10 -18 2009.3×10 -36 911209http:\\asadipour.kmu.ac.ir...46 slides

38. Gibbs Free Energy & equilibrium 1.If  G is negative, the forward reaction is spontaneous. 2.If  G is 0, the system is at equilibrium. 3.If  G is positive, the reaction is spontaneous in the reverse direction. 38911209http:\\asadipour.kmu.ac.ir...46 slides

39. 39 The Temperature Dependence of Equilibrium Constants ΔG = ΔH - T·ΔS Divide by RT, then multiply by -1 911209http:\\asadipour.kmu.ac.ir...46 slides

40. 40 Notice that this is y = mx +b the equation for a straight line A plot of y = mx + b or ln K vs. 1/T Endothermic R. T  K  Exothermic R. T  K  911209http:\\asadipour.kmu.ac.ir...46 slides

41. 41 If we have two different Temperatures and K’s ΔH and ΔS are constant over the temperature range 911209http:\\asadipour.kmu.ac.ir...46 slides

42. 42 The reaction 2 Al 3 Cl 9 (g) → 3 Al 2 Cl 6 (g) Has an equilibrium constant of 8.8X10 3 at 443K and a ΔH r °= 39.8 kJmol -1 at 443K. Estimate the equilibrium constant at a temperature of 600K. 911209http:\\asadipour.kmu.ac.ir...46 slides

43. 43911209http:\\asadipour.kmu.ac.ir...46 slides

44. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. 44911209http:\\asadipour.kmu.ac.ir...46 slides

45. Standard Entropies Standard entropies tend to increase with increasing molar mass. MW  S  45911209http:\\asadipour.kmu.ac.ir...46 slides

46. Benzene, C 6 H 6, (at 1 atm),boils at 80.1°C and ΔH vap = 30.8 kJ –A) Calculate ΔS vap for 1 mole of benzene at 60°C and pressure = 1 atm. 46 ΔG vap =ΔH vap -TΔS vap at the boiling point, ΔG vap = 0 ΔH vap = T b ΔS vap 911209http:\\asadipour.kmu.ac.ir...46 slides B) Does benzene spontaneously boil at 60°C?