1. CHE 116 No. 1 Chapter Nineteen Copyright © Tyna L. Meeks 2001 - 2002 All Rights Reserved

2. CHE 116 No. 2 Spontaneous Processes Understanding and designing chemical reactions: How rapidly does the reaction proceed? - reaction rates - controlled by a factor related to energy - the lower the activation energy, the faster the reaction proceeds

3. CHE 116 No. 3 Spontaneous Processes How far toward completion will the reaction go? How far toward completion will the reaction go? - equilibrium constants - depends on rates of forward and reverse reactions - equilibrium should also be dependent on energy in some way due to dependence on reaction rates

4. CHE 116 No. 4 Spontaneous Processes Chemical thermodynamics is the relationship between equilibrium and energy. First Law of Thermodynamics: for a reaction that occurs at constant pressure, the enthalpy change equals the heat transferred between the system and its surroundings Energy is conserved!! First Law of Thermodynamics: for a reaction that occurs at constant pressure, the enthalpy change equals the heat transferred between the system and its surroundings Energy is conserved!! * Enthalpy is important in helping us determine if a reaction will proceed! * Enthalpy is important in helping us determine if a reaction will proceed!

5. CHE 116 No. 5 Spontaneous Processes Spontaneous Processes: 3energy is neither created nor destroyed in any process, energy can only be transferred or converted from one form to another   E = q + w 3a spontaneous process occurs without any outside intervention as energy is conserved

6. CHE 116 No. 6 Spontaneous Processes

7. CHE 116 No. 7 Spontaneous Processes Spontaneous Processes: 3temperature is going to effect the spontaneity of a process 3if discussing a phase change, at the temperature of the phase change, the phases compete for spontaniety, and neither is said to win over the other

8. CHE 116 No. 8 Spontaneous Processes Sample exercise: Under 1 atm pressure, CO 2 (s) sublimes at -78°C. Is the transformation of CO 2 (s) to CO 2 (g) a spontaneous process at -100°C?

9. CHE 116 No. 9 Spontaneous Processes Sample exercise: Under 1 atm pressure, CO 2 (s) sublimes at -78°C. Is the transformation of CO 2 (s) to CO 2 (g) a spontaneous process at -100°C? *No, if the temp had been higher it would change phase spontaneously, but lower than the sublimation point favors the reverse, so the solid remains a solid.

10. CHE 116 No. 10 Spontaneous Processes Reversible and Irreversible Processes: 3State functions: define a state and do not depend upon the pathway 3temperature 3internal energy 3enthalpy

11. CHE 116 No. 11 Spontaneous Processes Reversible and Irreversible Processes: 3Reversible processes is a unique way for a system to change its state, than go back to its state by following the exact same path but in the opposite direction 3phase changes at constant temp 3only one specific value of q (heat) 3system in equilibrium

12. CHE 116 No. 12 Spontaneous Processes Reversible and Irreversible Processes: 3Irreversible processes cannot be simply restored to their original state using the same path, it may be forced to go back, but by a different pathway 3phase changes at different temps 3two q values need to be established q forward and q reverse 3any spontaneous reaction

13. CHE 116 No. 13 Spontaneous Processes Thermodynamics can tell us 3direction of reaction 3extent of reaction NOT 3speed of reaction

14. CHE 116 No. 14 Entropy and the 2nd Law Spontaniety depends upon two factors Enthalpy (  H): heat of reaction 3exothermic normally spontaneous 3endothermic normally NOT spontaneous Entropy (  S): disorder of the system 3natural law indicates reactions go in the direction that leads to more disorder

15. CHE 116 No. 15 Entropy and the 2nd Law The Spontaneous Expansion of a Gas: When the stopcock is opened, the gas will spontaneously flow to fill the empty half, but it WILL NOT flow backward without work being done on the system.

16. CHE 116 No. 16 Entropy and the 2nd Law The Spontaneous Expansion of a Gas: Gas expands because of the tendency for the molecules to ‘spread out’ among the different arrangements that they can take.

17. CHE 116 No. 17 Entropy and the 2nd Law Entropy: measurement of randomness or chaos 3melting ice 3dissolving salts 3The more disordered or random a system, the larger its entropy

18. CHE 116 No. 18 Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: a) CO 2 (s)  CO 2 (g)

19. CHE 116 No. 19 Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: a) CO 2 (s)  CO 2 (g) Entropy increases

20. CHE 116 No. 20 Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: b) CaO(s) + CO 2 (g)  CaCO 3 (s)

21. CHE 116 No. 21 Entropy and the 2nd Law Sample exercise: Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system: b) CaO(s) + CO 2 (g)  CaCO 3 (s) Entropy decreases

22. CHE 116 No. 22 Entropy and the 2nd Law Entropy: measurement of randomness or chaos 3state function 3for a process that occurs at constant temperature, the entropy change is dependent on the heat transferred during the reverse of the reaction (q rev )  S = q rev /T

23. CHE 116 No. 23 Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?

24. CHE 116 No. 24 Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?  S = q rev /T

25. CHE 116 No. 25 Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?  S = q rev /T q rev =  vap = 38.56kJ/mol = 38.56kJ/mol T = 351.45 K

26. CHE 116 No. 26 Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?  S = q rev /T 38.56kJ 1000 J 1 mol mol 1 kJ 46 g T = 351.45 K

27. CHE 116 No. 27 Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?  S = q rev /T 838.26 J/g * 25.8 g = -21627 J T = 351.45 K

28. CHE 116 No. 28 Entropy and the 2nd Law Sample exercise: The normal boiling point of ethanol, C 2 H 5 OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of C 2 H 5 OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?  S = q rev /T =-21627 J/351.45 K = -61.5 J/K = -61.5 J/K

29. CHE 116 No. 29 Entropy and the 2nd Law Second law of Thermodynamics: In any reversible process,  S univ = 0. In any irreversible reaction,  S univ >0.  S univ =  S sys +  S surr  S univ =  S sys +  S surr In an isolated system, just the entropy of the system is considered.

30. CHE 116 No. 30 Molecular Interpretation On the microscopic level, the number of gas molecules can be directly related to the amount of entropy in a system. 3The more gas molecules present, the higher the entropy value 3a phase change that increases the number of gas molecules would increase entropy 3a phase change that decreases the number of gas molecules would decrease entropy

31. CHE 116 No. 31 Molecular Interpretation Three moles of gas combine to form two moles of gas, thus decreasing the number of molecules.

32. CHE 116 No. 32 Molecular Interpretation Degrees of freedom 3creating new bonds decreases the freedom of movement atoms may have had. 33 degrees 3motion in one direction, translational movement 3vibrational movement 3spinning, rotational movement

33. CHE 116 No. 33 Molecular Interpretation Figure 19.12

34. CHE 116 No. 34 Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of H 2 (g) at STP or 1 mol of H 2 (g) at 100°C and 0.5 atm.

35. CHE 116 No. 35 Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of H 2 O(s) at 0°C or 1 mol of H 2 O(l) at 25°C.

36. CHE 116 No. 36 Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of H 2 (g) at STP or 1 mol of SO 2 (g) at STP.

37. CHE 116 No. 37 Molecular Interpretation Sample exercise: Choose the substance with the greatest entropy in each case: 1 mol of N 2 O 4 (g) at STP or 2 mol of NO 2 (g) at STP.

38. CHE 116 No. 38 Molecular Interpretation Sample exercise: Predict whether is  S is positive or negative in each of the following processes: HCl(g) + NH 3 (g)  NH 4 Cl(s)

39. CHE 116 No. 39 Molecular Interpretation Sample exercise: Predict whether is  S is positive or negative in each of the following processes: 2SO 2 (g) + O 2 (g)  2SO 3 (s)

40. CHE 116 No. 40 Molecular Interpretation Sample exercise: Predict whether is  S is positive or negative in each of the following processes: cooling of nitrogen gas from 20°C to -50°C

41. CHE 116 No. 41 Calculation of Entropy Changes Entropy Calculations: 3no easy method for measuring entropy 3experimental measurements on the variation of heat capacity with temperature can give a value known as absolute entropy 3zero point of reference for perfect crystalline solids 3tabulated as molar quantities, J/mol-K

42. CHE 116 No. 42 Calculation of Entropy Changes Entropy differs from enthalpy 3standard molar entropies are not 0 3entropies of gases are greater than those of liquids and solids 3entropies increase with molar mass 3entropies increase with number of atoms in formula   S° = nS(products) - mS(reactants)

43. CHE 116 No. 43 Gibbs Free Energy Spontaneity involves both thermodynamic concepts: entropy and enthalpy  G =  H – T  S G = Gibbs Free Energy H = Enthalpy T = Temperature (K) S = Entropy

44. CHE 116 No. 44 Gibbs Free Energy Gibbs Free Enegry If  G is negative, the reaction is spontaneous and proceeds in the forward direction If  G is negative, the reaction is spontaneous and proceeds in the forward direction If  G is zero, the reaction is at equilibrium If  G is zero, the reaction is at equilibrium If  G is positive, the reaction is nonspontaneous and proceeds in the reverse direction If  G is positive, the reaction is nonspontaneous and proceeds in the reverse direction

45. CHE 116 No. 45 Gibbs Free Energy Gibbs Free Enegry l State function l Table 19.3

46. CHE 116 No. 46 Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate  G  at 298 K for the combustion of methane: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)

47. CHE 116 No. 47 Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate  G  at 298 K for the combustion of methane: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) -50.8 0 -394.4 2(-228.57) -50.8 0 -394.4 2(-228.57)

48. CHE 116 No. 48 Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate  G  at 298 K for the combustion of methane: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) -50.8 0 -394.4 2(-228.57) -50.8 0 -394.4 2(-228.57)  G  = products - reactants

49. CHE 116 No. 49 Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate  G  at 298 K for the combustion of methane: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) -50.8 0 -394.4 2(-228.57) -50.8 0 -394.4 2(-228.57)  G  = products – reactants (-394.4 + 2(-228.57)) – (-50.8 + 0) (-394.4 + 2(-228.57)) – (-50.8 + 0)

50. CHE 116 No. 50 Gibbs Free Energy Sample exercise: By using data from Appendix C, Calculate  G  at 298 K for the combustion of methane: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) -50.8 0 -394.4 2(-228.57) -50.8 0 -394.4 2(-228.57)  G  = products – reactants (-394.4 + 2(-228.57)) – (-50.8 + 0) (-394.4 + 2(-228.57)) – (-50.8 + 0) -800.7 -800.7

51. CHE 116 No. 51 Gibbs Free Energy Sample exercise: Consider the combustion of propane to form CO 2 (g) and H 2 O(g) at 298K. Would you expect  G  to be more negative or less negative than  H  ?

52. CHE 116 No. 52 Gibbs Free Energy Sample exercise: Consider the combustion of propane to form CO 2 (g) and H 2 O(g) at 298K. Would you expect  G  to be more negative or less negative than  H  ? more negative, using Gibbs Free Enegry formula, more moles of gas being produced would be increasing entropy, +  S, more negative, using Gibbs Free Enegry formula, more moles of gas being produced would be increasing entropy, +  S,

53. CHE 116 No. 53 Free Energy and Temperature Table 19.4

54. CHE 116 No. 54 Free Energy and Temperature Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate  H  and  S  at 298 K for the following reaction: 2SO 2 (g) + O 2 (g)  2SO 3 (g)

55. CHE 116 No. 55 Free Energy and Temperature Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate  H  and  S  at 298 K for the following reaction: 2SO 2 (g) + O 2 (g)  2SO 3 (g) 2(-296.9) 0 2(-395.2) 2(-296.9) 0 2(-395.2)  H = 2(-395.2) - 2(-296.9) = -196.6 kJ

56. CHE 116 No. 56 Free Energy and Temperature Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix C, calculate  H  and  S  at 298 K for the following reaction: 2SO 2 (g) + O 2 (g)  2SO 3 (g) 2(248.5) 205.0 2(256.2) 2(248.5) 205.0 2(256.2)  S = 2(256.2) – (2(248.5)+205.0) = -189.6 J

57. CHE 116 No. 57 Free Energy and Temperature Sample exercise: Using the values obtained, estimate  G  at 400 K  G =  H – T  S  G =  H – T  S

58. CHE 116 No. 58 Free Energy and Temperature Sample exercise: Using the values obtained, estimate  G  at 400 K  G =  H – T  S  G =  H – T  S = -196.6 kJ – 400(-0.1896 kJ) = -196.6 kJ – 400(-0.1896 kJ) = -120.8 = -120.8

59. CHE 116 No. 59 Free Energy and the Equilibrium Constant 2 other important ways free energy is a powerful tool Tabulate free energy under nonstandard conditions Tabulate free energy under nonstandard conditions Directly relate free energy to equilibrium constants Directly relate free energy to equilibrium constants

60. CHE 116 No. 60 Free Energy and the Equilibrium Constant Tabulate free energy under nonstandard conditions  G =  G° + RTlnQ  G =  G° + RTlnQ R = 8.314 J/mol-K T = temperature K Q = reaction quotient

61. CHE 116 No. 61 Free Energy and the Equilibrium Constant Sample exercise: Calculate  G at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N 2, 0.75 atm H 2, and 2.0 atm NH 3.  G =  G° + RTlnQ

62. CHE 116 No. 62 Free Energy and the Equilibrium Constant Sample exercise: Calculate  G at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N 2, 0.75 atm H 2, and 2.0 atm NH 3.  G =  G° + RTlnQ 2(-16.6) + 0.008314(298)ln(2.0 2 /0.50(0.75 3 )) 2(-16.6) + 0.008314(298)ln(2.0 2 /0.50(0.75 3 )) -26.0 kJ -26.0 kJ

63. CHE 116 No. 63 Free Energy and the Equilibrium Constant At equilibrium,  G is equal to 0 so…  G° = -RTlnK eq  G° negative: K > 1  G° negative: K > 1  G° zero : K = 1  G° zero : K = 1  G° positive: K < 1  G° positive: K < 1 K eq = e -  G°/RT