1. Chemical Thermodynamics Chapter 17 Chemical Thermodynamics

2. Chemical Thermodynamics Spontaneous Processes Entropy Second Law of Thermodynamics Third Law of Thermodynamics Gibbs Free Energy Predicting Spontaneity Standard Enthalpies of Formation Gibbs Free Energies of Formation Free Energy Changes Contents

3. Chemical Thermodynamics Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer.

4. Chemical Thermodynamics First Law of Thermodynamics You will recall from Chapter 6 that 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.

5. Chemical Thermodynamics First law of thermodynamics, the law of conservation of energy, dictates the relationship between heat (q), work (w), and changes in internal energy ( ΔU ). ΔU = q + w Notice: assigning the correct signs to the quantities of heat and work.

6. Chemical Thermodynamics Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously separate from each other.

7. Chemical Thermodynamics Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

8. Chemical Thermodynamics Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C it is spontaneous for ice to melt. Below 0  C the reverse process is spontaneous.

9. Chemical Thermodynamics Goal of chemical thermodynamics: predicting which changes will be spontaneous.

10. Chemical Thermodynamics Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.

11. Chemical Thermodynamics Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system. Spontaneous processes are irreversible.

12. Chemical Thermodynamics Entropy Entropy (S) is a term coined by Rudolph Clausius in the 19th century. Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, qTqT

13. Chemical Thermodynamics Entropy Entropy can be thought of as a measure of the randomness of a system. It is related to the various modes of motion in molecules. The entropy of a system in a given state is a measure of the number of different microscopic states that correspond to a given macroscopic state.

14. Chemical Thermodynamics The more the particles and their positions, the more disordered the system is.

15. Chemical Thermodynamics Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tends to increase with increases in  Temperature.  Volume.  The number of independently moving molecules.

16. Chemical Thermodynamics Entropy Like total energy, U, and enthalpy, H, entropy is a state function. Therefore,  S = S final  S initial

17. Chemical Thermodynamics Entropy For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:  S = q rev T

18. Chemical Thermodynamics Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S (g) > S (l) > S (s)

19. Chemical Thermodynamics Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes.

20. Chemical Thermodynamics  S and Isolated Systems For an equilibrium process in an isolated system,  S = 0 For a spontaneous process in an isolated system,  S > 0

21. Chemical Thermodynamics Second Law of Thermodynamics In other words: For reversible processes:  S univ =  S system +  S surroundings = 0 For irreversible processes:  S univ =  S system +  S surroundings > 0

22. Chemical Thermodynamics Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.

23. Chemical Thermodynamics Entropy Changes In general, entropy increases when  Gases are formed from liquids and solids.  Liquids or solutions are formed from solids.  The number of gas molecules increases.  The number of moles increases.

24. Chemical Thermodynamics Solutions Generally, when a solid is dissolved in a solvent, entropy increases.

25. Chemical Thermodynamics Entropy and Disorder The entropy S of a system is a measure of the disorder or randomness in the system. Disorder can be defined as the number of equivalent ways of distributing the conserved matter and energy through the system. Example: a substance has higher entropy in the gaseous state than in the solid state.

26. Chemical Thermodynamics Entropy Increasing Processes The entropy is expected to increase for processes in which 1.liquids or solutions are formed from solids. 2.gases are formed from either liquids or solids. 3.the number of molecules of gas increases in going from reactants to products. 4.The number of degrees of freedom increases.

27. Chemical Thermodynamics Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

28. Chemical Thermodynamics Third Law of Thermodynamics The entropy of all pure crystalline substances approach zero as the temperature approaches absolute zero – since all disorder has been removed. as T  0 K, S 0  0 This defines the absolute entropy scale S 0.

29. Chemical Thermodynamics Standard Entropies The standard entropy of a substance S 0 is the entropy change required to heat 1 mole of the substance from 0 K to the temperature of 298 K. Standard molar entropies are used to calculate  S for reactions, just as  H f values are used to calculate  H for reactions. Note that S 0 for an element in its standard state is not zero – unlike the case for  H f

30. Chemical Thermodynamics Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass.

31. Chemical Thermodynamics Standard Entropies Larger and more complex molecules have greater entropies.

32. Chemical Thermodynamics Entropy Changes Entropy changes for a reaction can be estimated in a manner analogous to that by which  H is estimated:  S rxn ° =  n p S° (products) -  n r S° (reactants) where n and m are the coefficients in the balanced chemical equation.

33. Chemical Thermodynamics Sample problem to calculate  S rxn °: Use standard entropies to calculate  S rxn °for the reaction: 2SO 2 (g)+O 2 (g)= 2SO 3 (g) Solution: Equation: 2SO 2 (g)+O 2 (g)= 2SO 3 (g) S°, J/K/mol 248.12 205.03 256.72  S rxn ° =  n p S° (products) -  n r S° (reactants) =2× S° (SO 3 ) - 2× S° (SO 2 ) - × S° (O 2 ) =2 ×256.72-2 ×248.13-205.03 =-187.83 J/K/mol

34. Chemical Thermodynamics Entropy Exercise Predict whether the entropy change in each of the following reactions is positive or negative: a)CaCO 3 (s)  CaO(s) + CO 2 (g) b)N 2 (g) + 3H 2 (g)  2NH 3 (g) c)H 2 O(l)  H 2 O(g) d)Ag + (aq) + Cl - (aq)  AgCl(s) e)4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)

35. Chemical Thermodynamics Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process:  S surr =  q sys T At constant pressure, q sys is simply  H  for the system.

36. Chemical Thermodynamics Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore,  S universe =  S system +  S surroundings For spontaneous processes  S universe > 0

37. Chemical Thermodynamics Entropy Change in the Universe This becomes:  S universe =  S system + Multiplying both sides by  T,  T  S universe =  H system  T  S system  H system T

38. Chemical Thermodynamics Gibbs Free Energy  T  S universe is defined as the Gibbs free energy,  G. When  S universe is positive,  G is negative. Therefore, when  G is negative, a process is spontaneous.

39. Chemical Thermodynamics Criterion for Spontaneity According to the 2 nd Law, a reaction is spontaneous at constant pressure and temperature if and only if:  H system  T  S system <0 Or …  H - T  S < 0

40. Chemical Thermodynamics Gibbs Free Energy  These two factors are combined in the Gibbs free energy, defined:  G = H – TS or  G =  H - T  S  A reaction is spontaneous (under conditions of constant T and P) when  G < 0 Spontaneous reactions are favored by:   H < 0 (exothermic)   S > 0 (increasing entropy)

41. Chemical Thermodynamics Predicting Spontaneity 1.If a reaction has  G < 0 it is spontaneous (in the forward direction). 2.If a reaction has  G > 0 its reverse is spontaneous. 3.If a reaction has  G =0 then it is already at equilibrium.

42. Chemical Thermodynamics Gibbs Free Energies of Formation The Gibbs free energy of formation  G f  for a substance is defined in the same way as the enthalpy of formation (  H f  )..  G f  for a substance is the Gibbs free energy change when one mole of the substance is formed under standard conditions from its elements in their standard states.

43. Chemical Thermodynamics Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation,  G . f  G rxn  =  n p  G f  (products)   n r  G f  (reactants) where n p and n r are the stoichiometric coefficients.

44. Chemical Thermodynamics  G rxn  from  G f  Values Just as  S  of reaction can be calculated from standard entropy values S 0, and  H of reaction can be calculated from  H f  values, so  G rxn  of reaction can be calculated from  G f  values.  G rxn  =  n p  G f  (products)   n r  G f  (reactants)

45. Chemical Thermodynamics The difference between  G rxn  and  G rxn  G rxn  (or  G  ): is the free energy change that accompanies a change from reactants in their standard states to products in their standard states.  G rxn (or  G ): is the free energy change that accompanies a change from reactants in nonstandard states to products in nonstandard states.

46. Chemical Thermodynamics Example for calculation of  G°from  H°and  S°: Use standard heat of formation and standard entropies to calculate  G°for the reaction at 25 ℃ and 1atm partial pressure of each gas: 3H 2 (g)+N 2 (g)= 2NH 3 (g) Solution: (1)Calculation of  H° and  S°: Equation: 3H 2 (g)+N 2 (g)= 2NH 3 (g)  H f °, kJ/mol 0 0 -46.11 S°, J/K/mol 248.12 205.03 256.72  H°=  n p  H f °, (products) -  n r  H f °, (reactants) =2 ×(-46.11) - 3×0 - 2×0 =-92.22 kJ/mol

47. Chemical Thermodynamics  S rxn ° =  n p S°(products) -  n r S°(reactants) =2× S°(NH 3 ) - 3× S°(H 2 ) - 2× S°(N 2 ) =2 ×192.3 - 3 ×130.68 - 2 × 191.50 =-198.9 J/K/mol (2)Calculation of  G°:  G°=  H°-T  S°= -92.22 –(273.2+25) ×(-198.9 ×10 -3 ) =-32.91kJ/mol

48. Chemical Thermodynamics Example for  G 0 of Reaction from  G f 0 Calculate the standard Gibbs free energy of reaction for the combustion of methane CH 4. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  G f (CH 4,g ) = -50.8 kJ/mol  G f (CO 2,g) = -393.4 kJ/mol  G f (H 2 O,l) = -237.13 kJ/mol  G 0 = 2 ×  G f (H 2 O,l) +  G f (CO 2,g) -  G f (CH 4,g )  G 0 = -816.9 kJ

49. Chemical Thermodynamics  S ° and  H ° for Reactions where n p is the stoichiometric coefficient of product, n r is the stoichiometric coefficient of reactant.  S rxn ° =  n p S° (products) -  n r S° (reactants)  H rxn ° =  n p  H f °(product) -  n r  H f °( reactant)

50. Chemical Thermodynamics Free Energy Changes At temperatures other than 25°C,  G° =  H   T  S  How does  G  change with temperature?

51. Chemical Thermodynamics Free Energy and Temperature There are two parts to the free energy equation:   H  — the enthalpy term  T  S  — the entropy term The temperature dependence of free energy, then comes from the entropy term.

52. Chemical Thermodynamics Example for  G 0 from  H 0 and  S 0 Consider N 2 (g) + 3H 2 (g)  2NH 3 (g). Assume that  H 0 and  S 0 do not change much with temperature. Calculate  G 0 for the reaction at 500 K.  H 0 = -92.38 kJ/mol  S 0 = -198.3 J/K-mol  G 0 =  H 0 - T  S 0 = (-92.38) - (500) ×(-0.1983)  G 0 = +6.77 kJ

53. Chemical Thermodynamics S 0 : depend markedly on temperature.  S 0 : however, because increasing temperature increase the entropy of all substance,  S 0 often do not change greatly with temperature at ordinary temperature.  H 0 : are also often quite constant as temperature changes because of the same formation and cleavage in a certain reaction.

54. Chemical Thermodynamics Predicting Spontaneity Two factors determine the spontaneity of a chemical or physical change:  Enthalpy change  H  Entropy change  S  H < 0 (exothermic) favors the process.  S > 0 (more randomness) favors the process.

55. Chemical Thermodynamics  H,  S and Spontaneity There are four possible combinations of positive and negative  H and  S: 1.  H 0   G < 0 : spontaneous at any temperature 2.  H > 0 and  S 0 : not spontaneous at any temperature 3.  H > 0 and  S > 0  favored at high T 4.  H < 0 and  S < 0  favored at low T

56. Chemical Thermodynamics Temperature Dependence Since  G=0 at equilibrium, a process will reach equilibrium when  H = T  S or at the temperature T =  H/  S. If a process is non spontaneous at T  H/  S and vice versa.

57. Chemical Thermodynamics Free Energy and Temperature

58. Chemical Thermodynamics Energy and Equilibrium Energy  reactants products equilibrium Q: reaction quotient; K: equilibrium constant Equilibrium: A system’s macroscopic properties do not change spontaneously. (V forward =V reverse ) Q < Kc means the reaction will go spontaneously in the forward direction. Q > Kc means the reaction will go spontaneously in the reverse direction.

59. Chemical Thermodynamics Gibbs Free Energy 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.

60. Chemical Thermodynamics Free Energy and Equilibrium Under any conditions, standard or nonstandard, the free energy change can be found this way:  G =  G  + RT lnQ (Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

61. Chemical Thermodynamics Free Energy and Equilibrium At equilibrium, Q = K, and  G = 0. The equation becomes 0 =  G  + RT lnK Rearranging, this becomes  G  =  RT lnK or, K = e  G  /RT

62. What's Free about Free Energy The change in the Gibbs free energy for a process is the maximum amount of useful work that can be done by the system at constant temperature and pressure.

63. CaCO 3 Example At what temperature will CaCO 3 (s) just begin to decompose to CaO(s) and CO 2 (g) under standard conditions? CaCO 3 (s)  CaO(s) + CO 2 (g)  H 0 = +178.3 kJ  S 0 = +159.0 J/K = 0.159 kJ/K  G 0 = 0 at equilibrium.  H 0 - T  S 0 = 0  T =  H 0 /  S 0 = 1121 K ∴ When T>1121K, the reaction is spontaneous.

64. K from  G Example Consider N 2 (g) + 3H 2 (g)  2NH 3 (g). Calculate the equilibrium constant at 500 K.  H 0 = -92.38 kJ/mol  S 0 = -198.3 J/K-mol  G 0 =  H 0 - T  S 0 = (-92.38) – (500)(-0.1983)  G 0 = +6.77 kJ K = exp(-  G 0 / RT) = 0.196 (K = e  G  /RT )