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 CO 2 or gaseous CO 2 ? Solid CO 2 or gaseous CO 2 ? N 2 gas at 1 atm or N 2 gas at 0.01 atm? N 2 gas at 1 atm or N 2 gas at 0.01 atm? Predict the sign of  S Solid sugar is added to water Solid sugar is added to water Iodine vapor condenses on cold surface to form crystals Iodine vapor condenses on cold surface to form crystals

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

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

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

9. H 2 O(l)  H 2 O(g) ∆S sys has ____ sign b/c of the increase in # of positions ∆S surr is determined mostly by heat flow Vaporization is _______________ so it removes heat from the surroundings So ____________ random motion of surroundings ____________ ∆S surr

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

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

12. Summary

13. Example A process has a  H of +22 kJ and a  S of -13 J/K. At which temperatures is the process spontaneous? if there is no subscript,  S =  S sys if there is no subscript,  S =  S sys  S univ ___ 0 to be spontaneous  S univ ___ 0 to be spontaneous  S sys +  S surr > 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?  S sys = 213 J/K and  H = 71.8 kJ/mol K  S sys = 213 J/K and  H = 71.8 kJ/mol K at the boiling point, the vaporization begins to be spontaneous at the boiling point, the vaporization begins to be spontaneous  S univ = 0 to be at bpt   S sys +  S surr = 0  S univ = 0 to be at bpt   S sys +  S surr = 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 – TSdefinition of G for constant T

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

18. Example H 2 O(s)  H 2 O(l) If ∆H°=6.03x10 3 J/mol and ∆S°=22.1 J/mol. K and the temperature is at -10°C, is it spontaneous? If ∆H°=6.03x10 3 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 ∆S univ is positive Check to see if the ∆S univ is positive How can we solve for ∆S surr ? How can we solve for ∆S surr ? use ∆S surr = - ∆H°/T use ∆S surr = - ∆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 ∆S univ ° is 0, will ∆G° always be 0 too? How do the signs of ∆S univ ° 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 In opposition: +∆S and +∆H OR –∆S and -∆H Exothermic direction is spontaneous at low T Exothermic direction is spontaneous at low T

24. Example 2 At what T is this reaction spontaneous, at 1 atm of pressure? Br 2 (l)  Br 2 (g) (aka Boiling) Br 2 (l)  Br 2 (g) (aka Boiling) ∆H=31.0 kJ/mol, ∆S=93.0 J/molK ∆H=31.0 kJ/mol, ∆S=93.0 J/molK Looking for boiling point: ∆G=0= equilibrium 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 N 2 (g) + 3H 2 (g)  2NH 3 (g) N 2 (g) + 3H 2 (g)  2NH 3 (g) Entropy increases/decreases Entropy increases/decreases Compare using # molecules in higher entropy states 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Entropy increases/decreases Entropy increases/decreases

27. Example Predict the sign of ∆S for the following reactions: CaCO 3 (s)  CaO(s) + CO 2 (g) 2SO 2 (g) + O 2 (g)  2SO 3 (g)

28. Assigning S values Normally deal with changes in S but can assign actual S values to substances 3 rd law of thermodynamics: _______ for a perfect crystal at 0 K A standard value to help us set S values 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) + 3O 2 (g)  2SO 2 (g) + 2NiO(s) 2NiS(s) + 3O 2 (g)  2SO 2 (g) + 2NiO(s) 53 205 248 38 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 Can calculate ∆ H ° and ∆ S ° using appendix and products – reactants Then put in the equation to find ∆G° Then put in the equation to find ∆G°____________________ Rearrange equations to get the goal equation Rearrange equations to get the goal equation Add the ∆G° values for the equations 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 ∆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 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] aka [Products/Reactants]

36. Example Find the ∆G for the following reaction at 25°C and with CO at 5.0 atm and H 2 at 3.0 atm. CO(g) + 2H 2 (g)  CH 3 OH(l) must find ∆G° first using product and reactants values from appendix using product and reactants values from appendix calculate Q from pressure

37. Example 1 CO(g) + 2H 2 (g)  CH 3 OH(l) ∆G°= ∆G= ∆G° + RT ln Q

38. What does ∆G mean? a negative ∆G means 1. _ 2. _

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 N 2 (g) + 3H 2 (g)  2NH 3 (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 p NH3 = 1.00 atm, P N2 = 1.47, P H2 = 0.0100 ∆G = ∆G° + RT ln K ∆G =

43. Example 2 p NH3 = 1.00 atm, P N2 = 1.00, P H2 = 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 w max = ∆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