1. 1940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

2. Thermodynamics (Mark=2) 2940701http:\\academicstaff.kmu.ac.ir\alia sadipour...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 3940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

4. –Kinetic: –predicts speed (rate). –Thermodynamic: – predicts direction. Spontaneous Processes –without outside intervention –maybe slowly –C diamond → C grafit 4940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 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 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...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. 6940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

7. First Law of Thermodynamics E of adiabatic system is constant & incomputable.  E=E f -E i E f =E i +q-W E f -E i =q-W  E= q-W  E= state function 7940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

8. First Law of Thermodynamics  E= q-W 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 8940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

9. First Law of Thermodynamics R=Gases constant= 0.082 L.atm/mol.Kº R=Gases constant= 8.314 J/mol.Kº 9940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 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? 10940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

11. Summary of Entropy Entropy is the degree of randomness or disorder in a system S solid < S liquid < S gas The Entropy of all substances is positive 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 11940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 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 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 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 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

14. 14940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

15. 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. 15940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

16.  S system +  S surroundings =  S universe 16940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

17. Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial 17940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

18. 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 18940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

19. Free Energy and Chemical Reactions ΔG = ΔH - T·ΔS 19 W W q q ΔGΔG ΔGΔG ΔHΔH ΔHΔH TΔSTΔS TΔSTΔS Spontaneous reaction Ideal reverse cell Operating cell 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

20. Free Energy ΔG < 0Spontaneous ΔG = 0Equilibrium ΔG > 0Spontaneous Reverse Entropy ΔS univ > 0Spontaneous Forward ΔS univ = 0Equilibrium ΔS univ < 0Spontaneous Reverse 20940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

21. Effects of Temperature on ΔG° 3NO (g) → N 2 O (g) + NO 2 (g) 21940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

22. Spontaneity of reactions in different temperatures ΔG = ΔH - T·ΔS AB C D Case B ΔH° 0 ΔG = ΔH - T·ΔS ΔG < 0 spontaneous at all temp. Case C ΔH° > 0 ΔS° < 0 ΔG = ΔH - T·ΔS ΔG > 0 or non-spontaneous at all Temp. 22940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides Case A ΔH° > 0 ΔS° > 0 ΔG = ΔH - T·ΔS ΔG < 0 or spontaneous at high Temp. Case D ΔH° < 0 ΔS° < 0 ΔG = ΔH - T·ΔS ΔG < 0 or spontaneous at low Temp.

23. 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 23940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

24. Standard Entropies Standard entropies tend to increase with increasing molar mass. MW  S  24940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

25. 25940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

26. (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 26940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides mol= mol of reaction(total reaction=1 mol)

27. (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 ΔS rxn °= -153 JK -1 mol -1 2H 2 S(g) + 3O 2 (g) → 2SO 2 (g) + 2H 2 O(g) Solution: 27940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

28. 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 28940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

29. 29 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 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

30. 30 ΔG = ΔG° + RT ln Q Where Q is the reaction quotient a A + b B ⇋ c C + d D If Q = K ΔG = 0((equilibrium)) ΔG° = -RT ln K If Q > K ΔG = + the rxn shifts towards the reactant side If Q < K ΔG = - the rxn shifts toward the product side compare The Dependence of Free Energy & Pressure 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

31. 31 At Equilibrium conditions, ΔG = 0 ΔG° = -RT ln K a A + b B ⇋ c C + d D 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides NOTE: we can now calculate equilibrium constants (K) for reactions from standard ΔG f functions of formation

32. 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 32940701 ΔG rxn °= – 105 kJ mol -1 http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides Compare

33. 33 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 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

34. 34 ΔG = ΔG° + RT ln Q Where Q is the reaction quotient a A + b B ↔ c C + d D Spontaneous reactions 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

35. 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. 35940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides N 2 +3H 2 ⇋3NH 3 ΔG°=+5KJ Is the reaction spontaneoues at 25 0 C ???!!!  G is important not  G 0 ΔG = ΔG° + RT ln Q Standard C.

36. 36 The Temperature Dependence of Equilibrium Constants ΔG = ΔH - T·ΔS Divide by RT, then multiply by -1 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

37. 37 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 EXothermic R. T  K  Endothermic R. T  K  940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides T  K  T  K 

38. 38 If we have two different Temperatures and K’s ΔH and ΔS are constant over the temperature range 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides Log(K2/K1) = Ea/2.303R(T2-T1)/T1T2 Log(K2/K1) = ΔH /2.303R(T2-T1)/T1T2

39. 39 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. 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

40. 40940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

41. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. 41940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

42. Standard Entropies Standard entropies tend to increase with increasing molar mass. MW  S  42940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides

43. 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. 43 ΔG vap =ΔH vap -TΔS vap at the boiling point, ΔG vap = 0 ΔH vap = T b ΔS vap 940701http:\\academicstaff.kmu.ac.ir\alia sadipour...43 slides B) Does benzene spontaneously boil at 60°C?