Ch. 16: Spontaneity, Entropy, and Free Energy

1. Ch. 16: Spontaneity, Entropy, and Free Energy 16.1 Spontaneous Processes and Entropy

2. Spontaneous • Occurs without outside intervention • Thermodynamics tell us the _____________ not the speed • Thermodynamics only consider initial and final states, not pathway • Must use kinetics and thermodynamics to understand reaction completely • What makes a reaction spontaneous?

4. Entropy • S - measure of disorder or ____________ • the driving force for a spontaneous reaction is an increase in entropy • Natural tendency: low to high S • Entropy also describes the number of possible positions of a molecule

5. Example • Which has higher positional S? • Solid CO2 or gaseous CO2? • N2 gas at 1 atm or N2 gas at 0.01 atm? • Predict the sign of DS • Solid sugar is added to water • Iodine vapor condenses on cold surface to form crystals

6. Ch. 16: Spontaneity, Entropy, and Free Energy 16.2 Entropy and 2nd law of thermodynamics

7. 2nd Law of Thermodynamics • In any spontaneous reactions, there is always an increase in entropy of the universe ∆Suniv = ∆Ssys + ∆Ssurr • If ∆Ssys is negative, it can still be spont. as long as the ∆Ssurr is larger and positive • ∆Suniv > 0 : • ∆Suniv = 0 : • ∆Suniv < 0 :

8. Ch. 16: Spontaneity, Entropy, and Free Energy 16.3 The Effect of Temperature on Spontaneity

9. H2O(l)  H2O(g) • ∆Ssys has ____ sign b/c of the increase in # of positions • ∆Ssurr is determined mostly by heat flow • Vaporization is _______________ so it removes heat from the surroundings • So ____________ random motion of surroundings • ____________ ∆Ssurr

10. Temperature’s Effects • If the ∆Ssys and ∆Ssurr have different signs, the ______________ determines the ∆Suniv • For vaporization of water • Above 100°C, ∆Suniv is ____________ • Below 100°C, ∆Suniv is ____________ • Impact of the transfer of heat will be greater at lower temperatures

11. Determining ∆Ssurr • Sign of ∆Ssurr • depends on direction of heat flow • ∆Ssurr____ for exothermic reactions • ∆Ssurr____ for endothermic reactions • Magnitude of ∆Ssurr • Depends on temperature • Heat flow = ∆H at constant P • Very small at high T, increases as T decreases

12. Summary

13. Example • A process has a DH of +22 kJ and a DS of -13 J/K. At which temperatures is the process spontaneous? • if there is no subscript, DS = DSsys • DSuniv___ 0 to be spontaneous DSsys + DSsurr > 0

14. Example • For methanol, the enthalpy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 213 J/K. What is the normal boiling point of methanol? • DSsys= 213 J/K and DH =71.8 kJ/mol K • at the boiling point, the vaporization begins to be spontaneous • DSuniv = 0 to be at bpt DSsys + DSsurr = 0

15. Ch. 16: Spontaneity, Entropy, and Free Energy 16.4: Free Energy

16. Free Energy • G: helps you determine the temperature dependence of spontaneity • G ≡ H – TS definition of G for constant T

17. Free Energy and Entropy • So what sign would the ∆G of reaction with a + ∆Suniv have? So for a reaction at constant T and P:

18. Example • H2O(s)  H2O(l) • If ∆H°=6.03x103 J/mol and ∆S°=22.1 J/mol.K and the temperature is at -10°C, is it spontaneous?

19. Example 1 • What could be another way to check for spontaneity of a reaction? • Check to see if the ∆Suniv is positive • How can we solve for ∆Ssurr? • use ∆Ssurr= - ∆H°/T • At -10°C, is it spontaneous?

20. Example 1 • Is it spontaneous at 0°C?

21. Example 1 • Is it spontaneous at +10°C?

22. Example 1 When ∆Suniv° is 0, will ∆G° always be 0 too? How do the signs of ∆Suniv° and ∆G° relate?

23. Enthalpy and Entropy • When H and S are in opposition, the spontaneity depends on T • In opposition: +∆S and +∆H OR –∆S and -∆H • Exothermic direction is spontaneous at low T

24. Example 2 • At what T is this reaction spontaneous, at 1 atm of pressure? • Br2(l)  Br2(g) (aka Boiling) • ∆H=31.0 kJ/mol, ∆S=93.0 J/molK • Looking for boiling point: ∆G=0= equilibrium

25. Ch. 16: Spontaneity, Entropy, and Free Energy 16.5: Entropy Changes in Chemical Reactions

26. What about Chemical Changes? • We have been working with only physical changes so far… • Compare using # of independent states • N2(g) + 3H2(g)  2NH3(g) • Entropy increases/decreases • Compare using # molecules in higher entropy states • 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) • Entropy increases/decreases

27. Example • Predict the sign of ∆S for the following reactions: • CaCO3(s)  CaO(s) + CO2(g) • 2SO2(g) + O2(g)  2SO3(g)

28. Assigning S values • Normally deal with changes in S but can assign actual S values to substances • 3rd law of thermodynamics: _______ for a perfect crystal at 0 K • A standard value to help us set S values • Appendix 4 contains S° for common substances at 298 K and 1 atm • Since S is state function, can find ∆S by subtracting final - initial

29. Example 3 • Find the ∆S° at 25°C for: • 2NiS(s) + 3O2(g)  2SO2(g) + 2NiO(s) 53 205 248 38

30. Ch. 16: Spontaneity, Entropy, and Free Energy 16.6: Free Energy and Chemical Reactions

31. Standard Free Energy Change • ∆G°: change when reactants and products are both in standard states • Cannot measure directly but can calculate it from other measured values • Standard states: p. 260 chart • The more negative ∆G° is, the further a reaction’s ____________ position lies to the _________

32. 3 Ways to Calculate ∆G° • _________________________________ • Can calculate ∆ H ° and ∆ S ° using appendix and products – reactants • Then put in the equation to find ∆G° • ____________________ • Rearrange equations to get the goal equation • Add the ∆G° values for the equations • ___________ : standard free energy of formation

33. Ch. 16: Spontaneity, Entropy, and Free Energy 16.7: Free Energy and Pressure

34. Dependence on Pressure • ∆H is not dependent on the pressure of the system but ∆S is- Why? • ∆ G is dependent on both ∆ H and ∆ S so is also dependent on pressure • can calculate the ∆G of a substance at a pressure other than 1 atm using: • where • ∆G° is the change in free energy at 1 atm and ∆G is the change in free energy at the P • where R (_______________) and T is Kelvin

35. Dependence on Pressure • Can also determine the ∆G for a reaction using the reaction quotient (Q) • where R is gas law constant (8.314 J/molK) and Q is the reaction quotient in pressures (gases only) or concentrations • aka [Products/Reactants]

36. Example • Find the ∆G for the following reaction at 25°C and with CO at 5.0 atm and H2 at 3.0 atm. CO(g) + 2H2(g)  CH3OH(l) • must find ∆G° first • using product and reactants values from appendix • calculate Q from pressure

37. Example 1 CO(g) + 2H2(g)  CH3OH(l) • ∆G°= • ∆G= ∆G° + RT ln Q

38. What does ∆G mean? • a negative ∆G means • _ • _

39. Ch. 16: Spontaneity, Entropy, and Free Energy 16.8: Free Energy and Equilibrium

40. Equilibrium • no matter how much of the reactants or products are initially mixed, at a given set of conditions, the equilibrium position (K) will be the same • at equilibrium, ∆G= 0 and Q= K

41. Equilibrium • ∆G°=0: • K=1 • G°react= G°prod • ∆G°<0: • K>1 • G°react> G°prod • ∆G°>0: • K<1 • G°react< G°prod

42. Example 2 N2(g) + 3H2(g)  2NH3(g) • For the reaction above, ∆G° =-33.3 kJ/mol. For each of the mixtures below at 25°C, predict the direction in which the system will shift to reach equilibrium • pNH3 = 1.00 atm, PN2 = 1.47, PH2 = 0.0100 • ∆G = ∆G° + RT ln K • ∆G =

43. Example 2 • pNH3 = 1.00 atm, PN2 = 1.00, PH2 = 1.00 • ∆G = ∆G° + RT ln K • ∆G=

44. Temperature Dependence of K • A plot of ln K vs. 1/T is linear where • y = ln K and x = 1/T (in Kelvin) • m = -∆H°/R • b = ∆ S°/R

45. Ch. 16: Spontaneity, Entropy, and Free Energy 16.9: Free Energy and Work

46. Work • Can calculate the maximum amount of work that can be done by a process • wmax = ∆G • if the ∆G is +, then w is the amount of work expended to make a process occur • impossible to achieve maximum work because of wasted energy