1. Thermodynamics: Spontaneity, Entropy and Free Energy

2. Spontaneity A spontaneous process is one that occurs without outside intervention. Examples include: - a ball rolling downhill - ice melting at temperatures above 0 o C - gases expanding to fill their container - iron rusts in the presence of air and water - two gases mixing

3. Spontaneity Spontaneous processes can release energy (a ball rolling downhill), require energy (ice melting at temperatures above 0 o C), or involve no energy change at all (two gases mixing).

4. Spontaneity There are three factors that combine to predict spontaneity. They are: 1. Energy Change 2. Temperature 3. Entropy Change

5. Entropy A measure of randomness or disorder

6. Entropy Entropy, S, is a measure of randomness or disorder. The natural tendency of things is to tend toward greater disorder. This is because there are many ways (or positions) that lead to disorder, but very few that lead to an ordered state.

7. Entropy The driving force for a spontaneous process is an increase in the entropy of the universe.

8. Entropy

9. ΔS o and Phase Changes Gases have more entropy than liquids or solids.

10. ΔS o and Mixtures Mixtures have more entropy than pure substances.

11. Entropy Values of Common Substances

12. The 2 nd Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe.

13. The 2 nd Law of Thermodynamics Water spontaneously freezes at a temperature below 0 o C. Therefore, the process increases the entropy of the universe. The water molecules become much more ordered as they freeze, and experience a decrease in entropy. The process also releases heat, and this heat warms gaseous molecules in air, and increases the entropy of the surroundings.

14. The 2 nd Law of Thermodynamics Since the process is spontaneous below 0 o C, ΔS surr, which is positive, must be greater in magnitude than ΔS of the water molecules.

15. Δ S and Spontaneity

16. Spontaneity Entropy, temperature and heat flow all play a role in spontaneity. A thermodynamic quantity, the Gibbs Free Energy (G), combines these factors to predict the spontaneity of a process. ΔG = ΔH - TΔS

17. Spontaneity If a process releases heat (ΔH is negative) and has an increase in entropy (ΔS is positive), it will always be spontaneous. The value of ΔG for spontaneous processes is negative.

18. Spontaneity ΔG = ΔH - TΔS

19. Spontaneity and ΔG If ΔG is negative, the process is spontaneous (and the reverse process is non- spontaneous). If ΔG is positive, the process is non- spontaneous, and the reverse process is spontaneous. If ΔG = 0, the system is at equilibrium.

20. ΔGΔGΔGΔG Although ΔG can be used to predict in which direction a reaction will proceed, it does not predict the rate of the reaction. For example, the conversion of diamond to graphite has a ΔG o = -3 kJ, so diamonds should spontaneously change to graphite at standard conditions. However, kinetics shows that the reaction is extremely slow.

21. The Significance of ΔG ΔG represents the driving force for the reaction to proceed to equilibrium.

22. The Significance of ΔG If negative, the value of ΔG in KJ is the maximum possible useful work that can be obtained from a process or reaction at constant temperature and pressure. If positive, the value of ΔG in KJ is the minimum work that must be done to make the non-spontaneous process or reaction proceed.

23. Predicting the sign of ΔS o For many chemical reactions or physical changes, it is relatively easy to predict if the entropy of the system is increasing or decreasing. If a substance goes from a more ordered phase (solid) to a less ordered phase (liquid or gas), its entropy increases.

24. Predicting the sign of ΔS o For chemical reactions, it is sometimes possible to compare the randomness of products versus reactants. 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) The production of a gaseous product from a solid reactant will have a positive value of ΔS o.

25. Calculating Entropy Changes Since entropy is a measure of randomness, it is possible to calculate absolute entropy values. This is in contrast to enthalpy values, where we can only calculate changes in enthalpy. A perfect crystal at absolute zero has an entropy value (S) =0. All other substances have positive values of entropy due to some degree of disorder. A perfect crystal at absolute zero has an entropy value (S) =0. All other substances have positive values of entropy due to some degree of disorder.

26. Calculating Entropy Changes Fortunately, the entropy values of most common elements and compounds have been tabulated. Most thermodynamic tables, including the appendix in the textbook, include standard entropy values, S o.

27. Entropy Values of Common Substances

28. Calculating Entropy Changes For any chemical reaction, Δ S o reaction = Σmol prod S o products - Σmol react S o reactants The units of entropy are joules/K-mol.

29. Calculation of ∆G o ∆G o, the standard free energy change, can be calculated in several ways. ∆G o, the standard free energy change, can be calculated in several ways. ∆G o = ∆H o - T ∆S o It can be calculated directly, using the standard enthalpy change and entropy change for the process.

30. Calculation of ∆G o ∆G o = ∆H o - T ∆S o ∆G o = ∆H o - T ∆S o ∆H o is usually calculated by using standard enthalpies of formation, ∆H f o. ∆H o is usually calculated by using standard enthalpies of formation, ∆H f o. ∆H o rxn = Σmol prod ∆H o products - Σmol react ∆H o reactants

31. Calculation of ∆G o ∆G o = ∆H o - T ∆S o ∆G o = ∆H o - T ∆S o Once ∆H o and ∆S o have been calculated, the value of ∆G o can be calculated, using the temperature in Kelvins.

32. Calculation of ∆G o ∆G o can also be calculated by combining the free energy changes of related reactions. This is the same method used in Hess’ Law to calculate enthalpy changes. If the sum of the reactions gives the reaction of interest, the sum of the ∆G o values gives ∆G o for the reaction. ∆G o can also be calculated by combining the free energy changes of related reactions. This is the same method used in Hess’ Law to calculate enthalpy changes. If the sum of the reactions gives the reaction of interest, the sum of the ∆G o values gives ∆G o for the reaction.

33. Calculation of ∆G o Lastly, ∆G o can be calculated using standard free energies of formation, ∆G f o. Some tables of thermodynamic data, including the appendix of your textbook, include values of ∆G f o. Lastly, ∆G o can be calculated using standard free energies of formation, ∆G f o. Some tables of thermodynamic data, including the appendix of your textbook, include values of ∆G f o. ∆G rxn o = Σmol prod ∆G f o prod - Σmol react ∆G f o react

34. Calculation of ∆G o When calculating ∆G o from standard free energies of formation, keep in mind that ∆G f o for any element in its standard state is zero. As with enthalpies of formation, the formation reaction is the reaction of elements in their standard states to make compounds (or allotropes).

35. Calculation of ∆G o

36. Note the values of zero for nitrogen, hydrogen and graphite.

37. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Calculate ∆G o using the tables in the appendix of your textbook. Is the process spontaneous at this temperature? Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous?

38. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Calculate ∆G o using the tables in the appendix of your textbook. Is the process spontaneous at this temperature? Calculate ∆G o using the tables in the appendix of your textbook. Is the process spontaneous at this temperature? Calculation of ∆G rxn o will indicate spontaneity at 25 o C. It can be calculated using ∆G f o values or from ∆H f o and ∆S o values.

39. Calculation of ∆G o CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G rxn o = Σn prod ∆G f o prod - Σn react ∆G f o react ∆G rxn o = Σn prod ∆G f o prod - Σn react ∆G f o react

40. Calculation of ∆G o CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G rxn o =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)] ∆G rxn o =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)]

41. Calculation of ∆G o CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G rxn o =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)] = +130.4 kJ ∆G rxn o =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)] = +130.4 kJ

42. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Calculate ∆G o using the tables in the appendix of your textbook. Is the process spontaneous at this temperature? Calculate ∆G o using the tables in the appendix of your textbook. Is the process spontaneous at this temperature? Since ∆G rxn o =+130.4 kJ, the reaction is not spontaneous at 25 o C.

43. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous?

44. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? At 25 o C, ∆G rxn o is positive, and the reaction is not spontaneous in the forward direction.

45. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? Inspection of the reaction shows that it involves an increase in entropy due to production of a gas from a solid.

46. Spontaneity Problem Consider the reaction: Consider the reaction: CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. CaCO 3 (s) ↔CaO(s) + CO 2 (g) at 25 o C. Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? Is it spontaneous at all temperatures? If not, at what temperature does it become spontaneous? We can calculate the entropy change and the enthalpy change, and then determine the temperature at which spontaneity will occur.

47. CaCO 3 (s) ↔CaO(s) + CO 2 (g) Since ∆G o = ∆H o - T∆S o, and there is an increase in entropy, the reaction will become spontaneous at higher temperatures. To calculate ∆S o, use the thermodynamic tables in the appendix.

48. CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆S rxn o =[1mol(213.6J/K-mol)+1mol(39.7J/K-mol)] -[1mol(92.9J/K-mol)] = 160.4 J/K

49. CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o Since we know the value of ∆G o (+130.4 kJ) and ∆S o ( ∆H o at 25 o C. Since we know the value of ∆G o (+130.4 kJ) and ∆S o (160.4 J/K), we can calculate the value of ∆H o at 25 o C. 130.4 kJ = ∆H o –(298K) ( 130.4 kJ = ∆H o –(298K) (160.4 J/K) ∆H o = 130.4 kJ + (298K) (. ∆H o = 130.4 kJ + (298K) (.1604 kJ/K) ∆H o = + 178.2 kJ

50. CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o If we assume that the values of ∆H o and ∆S o don’t change much with temperature, we can estimate the temperature at which the reaction will become spontaneous.

51. CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o ∆G o is positive at lower temperatures, and will be negative at higher temperatures. Set ∆G o equal to zero, and solve for temperature. ∆H o - T∆S o 0 = ∆H o - T∆S o T = ∆H o ∆S o ∆S o

52. CaCO 3 (s) ↔CaO(s) + CO 2 (g) ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o ∆H o - T∆S o 0 = ∆H o - T∆S o T = ∆H o ∆S o ∆S o T = (178.2 kJ)/( T = (178.2 kJ)/(160.4 J/K)(10 -3 kJ/J) =1111K or 838 o C The reaction will be spontaneous in the forward direction at temperatures above 838 o C.

53. ∆G for Non-Standard Conditions The thermodynamic tables are for standard conditions. This includes having all reactants and products present initially at a temperature of 25 o C. All gases are at a pressure of 1 atm, and all solutions are 1 M.

54. ∆G for Non-Standard Conditions For non-standard temperature, concentrations or gas pressures: ∆G = ∆G o + RTlnQ Where R = 8.314 J/K-mol T is temperature in Kelvins Q is the reaction quotient

55. ∆G for Non-Standard Conditions For non-standard temperature, concentrations or gas pressures: ∆G = ∆G o + RTlnQ For Q, gas pressures are in atmospheres, and concentrations of solutions are in molarity, M.

56. ∆G o and Equilibrium A large negative value of ∆G o indicates that the forward reaction or process is spontaneous. That is, there is a large driving force for the forward reaction. This also means that the equilibrium constant for the reaction will be large.

57. ∆G o and Equilibrium A large positive value of ∆G o indicates that the reverse reaction or process is spontaneous. That is, there is a large driving force for the reverse reaction. This also means that the equilibrium constant for the reaction will be small. When a reaction or process is at equilibrium, ∆G o = zero.

58. ∆G o and Equilibrium

59. ∆G = ∆G o + RT lnQ At equilibrium, ∆G is equal to zero, and Q = K. 0 = ∆G o + RT lnK ∆G o = - RT lnK

60. ∆G o and Equilibrium Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: C (s, diamond) ↔ C (s, graphite)

61. ∆G o and Equilibrium Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: C (s, diamond) ↔ C (s, graphite) ∆G o = (1 mol) ∆G o f (graphite) - (1 mol) ∆G o f (diamond) = 0 -(1 mol)(2.900 kJ/mol) = -2.900 kJ The reaction is spontaneous at 25 o C.

62. ∆G o and Equilibrium Calculate, ∆G o and K at 25 o C for: Calculate, ∆G o and K at 25 o C for: C (s, diamond) ↔ C (s, graphite) ∆G o = -2.900 kJ ∆G o = -2.900 kJ = -RT ln K -2.900 kJ = -(8.314J/mol-K) (298.2K)ln K ln K = 1.170 K= e 1.170 = 3.22

63. C (s, diamond) ↔ C (s, graphite) The negative value of ∆G o and the equilibrium constant >1 suggest that diamonds can spontaneously react to form graphite. Although the reaction is thermodynamically favored, the rate constant is extremely small due to a huge activation energy. The disruption of the bonding in the diamond to form planar sp 2 hybridized carbon atoms is kinetically unfavorable.