1. Free Energy and Thermodynamics Perpetual motion is forbidden by the Laws of Thermodynamics 1

2. Chapter 16 Spontaneity, Entropy And Free Energy 1st Law of Thermodynamics Energy can be neither created nor destroyed. Energy of the universe is constant However not all energy is usable The 1 st Law is an energy bookkeeping system Helps answer questions: How much energy in involved? Does energy flow in or out of system? What form does the energy assume? 2

3. Spontaneous A reaction is spontaneous if it occurs without outside intervention Ex. Fe + H 2 O + O 2  Fe 2 O 3 (Rust) Does not mean fast We will be using Thermochemistry (Ch.6), Kinetics (Ch.12) & Thermodynamics (Ch. 16) to describe a reaction completely. Thermodynamics tells the direction not speed Thermodynamics compares initial & final states Kinetics describes pathway between reactant & product. 3

4. Formula Review 4

5. Spontaneity A process that is spontaneous in one direction will not be spontaneous in the other. Ex. If wood burns into CO 2 and H 2 O, Why can we not put CO 2 and H 2 O together and make wood? Processes will proceed without outside assistance If steel is exposed to air and moisture it rusts It’s independent of time The characteristic common in all spontaneous processes is an increase in the property called Entropy (S) If  S (universe) > 0, the reaction is spontaneous 5

6. Entropy Entropy is a measure of molecular randomness or disorder High Disorder = High Entropy Entropy is a thermodynamic function that describes the number of arrangements (positions) available to a system, it’s probability. Entropy is a measure of chaos in a system Entropy is a state function Driving force in all spontaneous processes is increase in Entropy (S) of the universe 6

7. Entropy Formal definition: Entropy (S) is a thermodynamic function that increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state. First expressed by Ludwig Boltzmann in early 1900 S = k ln W k is Boltzmann constant = gas constant (8.31) / Avogadro’s number = 1.38 X 10 -23 J/K W = number of energetically equivalent ways 7

8. 2 possible arrangements 50 % chance of finding the left empty Each configuration that gives a particular arrangement is called a microstate 1 MOLECULE OF GAS Probability Consider the number of arrangements for a system in the following evacuated bulbs 8

9. l 4 possible arrangements l 25% chance of finding the left empty l 50 % chance of them being evenly dispersed l 25% chance of finding the right empty 2 MOLECULES OF GAS See Table 16.1 p.753 Probability 9

10. Positional Entropy Positional Entropy (S) – depends upon the number of microstates a set of molecules can have in space 1.Solids have the fewest possible number of microstates in space, atoms are very ordered, crystal structures 2.Liquids have many more ways for molecules to be arranged than a solid 3.Gases have a huge number of positions possible S solid < S liquid << S gas A chemical system proceeds in a direction that increases the entropy of the universe. 10

11. Positional Entropy Choose the substance with the higher positional entropy: 1.CO 2(s) vs. CO 2(g) 2.N 2(g) at 1 atm vs. N 2(g) at 1.0 X 10 -2 atm As the change in entropy (∆S) becomes more positive the system/universe becomes more disorder. 11

12. Entropy Values @ 298K J/mol K SolidsLiquidsGases C (graph.) 5.7H2OH2O69.9H2H2 130.7 Ca41.4CH 3 OH126.8O2O2 205.1 MgO26.9Br 2 152.2Cl 2 223.0 NaCl72.1 12

13. “Entropy Ain’t What It Used To Be” 2nd Law of Thermodynamics: Entropy of the universe is always increasing If process is spontaneous Total entropy must increase ∆S (universe) is always increasing Entropy defined in terms of probability – Describes the number of arrangements available to a system existing in a given state – The most likely of the particles is the most random – Nature proceeds toward the system with the highest number of possible states 13

14. Predicting Entropy Changes Predict the sign of the entropy ∆S for each of the following processes. 1.Solid sugar is added to water to form a solution. 2.Iodine vapor condenses on a cold surface to form a crystal Remember: S solid < S liquid << S gas Use this formula to help determine ∆S ∆S = S (Final) – S (Initial) Which has a greater number of arrangement: Final state or Initial state 14

15. Entropy Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate. Remember 2nd Law of Thermodynamics, Entropy of universe is always increase  S universe =  S system +  S surroundings If  S univ process is spontaneous in forward direction If  s univ process is spontaneous in opposite direction If  s univ = 0, process has no tendency to occur = equilibrium 15

16. Exothermic processes (heat  out)  S surr is positive Endothermic processes (heat  )  S surr is negative Consider this process: H 2 O (l)  H 2 O (g) Is it spontaneous? i.e. Is  s univ positive)? How much volume will 1 mole of liquid and gas occupy? (Stoichiometry)  S sys is positive (liquid  gas,  entropy)  S surr is negative (vaporization is an endothermic process)  s sys = s products -  s reactants  s univ =  s sys +  s surr Effect of Temperature on Spontaneity 16

17. Entropy Changes In The Surroundings Energy changes in the surroundings are primarily determined by flow of heat. i.e the sign of ∆S surr depends on direction of heat flow. – An exothermic process is favored because by giving up heat, the entropy of the surroundings increases. – Summary 17

18. Entropy Changes In The Surroundings How much of an impact the transfer of heat is on the surroundings depends on how high the temperature is. If the transfer occurs at low temperature,  S surr will be greater. Magnitude of  S surr depends directly on quantity of heat transferred Magnitude of  S surr depends inversely on temperature 18

19. Enthalpy H Enthalpy (H) is the internal energy (E) of a system plus the pressure times volume (PV) H=E+PV  H = a measure of the total change of internal energy of a system.  H heat flow into system (endothermic)  H heat flow into system (exothermic) Entropy (S) and Enthalpy (H): two factors that determine spontaneity Nature will always transfer thermodynamic energy from a higher to a lower temperature 19

20. Determining  S surr In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore @ 25 o C Iron is used to reduce antimony in sulfide ores. Sb 2 S 3(s) + 3Fe (s)  2Sb (s) + 3FeS (s)  H= -125kJ Carbon is used as the reducing agent for oxide ores. Sb 4 O 6(s) + 6C (s)  4Sb (s) + 6CO (g)  H= 778kJ Label RedOx Values & Calculate  S surr for each reaction 20

21. Spontaneous or Not?  S sys  S surr  S univ Spontaneous? +++ Yes --- Yes in Reverse +- ? Yes If  S sys >  S surr -+ ? Yes If  S sys <  S surr 21

22. Gibb's Free Energy Gibb's Free Energy is defined as: G=H-TS G=Free Energy H=Enthalpy T=Temp.(K) S=Entropy  G=  H-T  S ( @ constant temperature) (  H &  S without subscripts means system) Study the derivation of formulas on p. 760 If  G < 0 @ constant T and P, the Process is spontaneous in which the free energy is decreasing -  G means +  S 22

23. Let’s Check For the reaction H 2 O(s)  H 2 O(l)  S o = 22.1 J/K mol  H o =6030 J/mol Calculate  G at 10 o C, 0 o C, & -10 o C (convert to K) Superscript o means substance is in standard state Use equation  G 0 =  H 0 -T  S 0 Spontaneity can be predicted from the sign of  H and  S. 23

24. At What Temperature Is The Following Process Spontaneous? Br 2(l)  Br 2(g) ∆H 0 = 31.0 kJ/mol ∆S 0 = 93.0 J/K What is the normal boiling point for Br 2(l) 24

25. At What Temperature Is The Following Process Spontaneous? CH 3 OH (l)  CH 3 OH (g) 1.Get H o and S o values from Appendix 4 2.Caution: Make sure your units are compatible 3.Plug values into formula 25

26. Summary  G=  H-T  S SS HH Spontaneous ? +- At all temp. ++ At high temp. “entropy driven” -- At low temp. “enthalpy driven” -+ Not at any temp. Reverse at all temp. 26

27. Predict the sign of ∆S for each of the following reactions. 1.Thermal decomposition of solid calcium carbonate CaCO 3(s)  CaO (s) + CO 2(g) 2.The oxidation of SO 2 in air 2SO 2(g) + O 2(g) → 2SO 3(g) In general, when a reaction involves gaseous molecules, the change in positional entropy is dominated by the relative number of molecules of reactants and products. 27

28. Third Law of Thermodynamics Entropy (S) Of A Pure Crystal At 0K = 0 Walther Hermann Nernst (1864-1941) Gives us a starting point All others must be >0 Standard Entropies values S o (@298 K & 1 atm) of substances are listed Appendix 4 Entropy is a state function of a system and ∆ S of a given reaction can be calculated using the formula: 28

29. Calculate ∆ S o Predict sign of ∆ S o Calculate ∆ S o at 25 o C 2NiS (s) + 3O 2(g) → 2SO 2(g) + 2 NiO (s) Substance S o (J/Kmol) SO 2(g) 248 NiO (s) 38 O 2(g) 205 NiS (s) 53 29

30. Calculate ∆ S o Calculate ∆ S o for the reduction of aluminum oxide by hydrogen gas Al 2 O 3(s) + 3H 2(g) → 2Al (s) + 3H 2 O (g) SubstanceS o (J/Kmol) Al 2 O 3(s) 51 H 2(g) 131 Al (s) 28 H 2 O (g) 189 30

31. Free Energy and Chemical Reactions  G o = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state.  G o can’t be measured directly, can be calculated from other measurements.  G o =  H o -T  S o Can calculate  H o using enthalpy of formation f equation p.246 31

32. Calculating  H o,  S o, &  G o Consider the reaction: 2SO 2(g) + O 2(g)  2SO 3(g) Carried out at 25 o C & 1 atm, Use the following data: Substance  H f o (kJ/mol) S o (J/Kmol) SO 2 -297248 SO 3 -396257 O2O2 0205 32

33. Hess’s Law “More than one way to skin a cat” Because ∆G is a state function you can obtain ∆G for reactions using Hess’s Law 2CO (g) + O 2(g)  2CO 2(g) Use a couple of related equation and manipulate them to get ∆G o 2CH 4(g) + 3O 2(g)  2CO (g) + 4H 2 O (g) ∆G o = -1088kJ CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) ∆G o = -801kJ See Ch 6.3 p. 242 for review on Hess’s Law 33

34. Calculate  G o Using the following data at 25 o C C diamond(s) + O 2(g)  CO 2(g)  G o = -397 kJ C graphite(s) + O 2(g)  CO 2(g)  G o = -394 kJ Calculate  G o for the reaction: C diamond(s)  C graphite(s) 34

35. Free Energy in Reactions Standard free energy of formation (  G o f ) is change in free energy that accompanies the formation of 1 mol of that substance from reactants and products in their standard state  G o f The standard free energy of formation for any element in its standard state is 0, ( see table on p.A19) Use the following formula: 35

36. Calculate  G o Methanol is a high-octane fuel used in high- performance racing engines. Calculate  G o for the reaction, given the  G f o below: 2CH 3 OH (g) + 3O 2(g)  2CO 2(g) + 4H 2 O (g) Substance GfoGfo CH 3 OH (g) -163 O 2(g) 0 CO 2(g) -394 H 2 O (g) -229 36

37. Free Energy is Dependent on Pressure Entropy depends on pressure because it effects volume. S large volume > S small volume Because volume and pressure are inversely related S low pressure > S high pressure Study the Derivation of  G =  G o +RTln(Q) 37

38. Calculate  G One method for synthesizing methanol CH 3 OH (l) involves reacting carbon monoxide and hydrogen gas @25 o C. Calculate  G for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to methanol. CO (g) + H 2(g) → CH 3 OH (l) 1 st calc.  G o Get  G f o values p. A19 & formula p. 769 2nd calc.  G Using  G =  G o +RTln(Q) 38

39. How far? As a reaction proceeds toward Equilibrium, the  G changes until it reaches zero which is the Equilibrium point.  G = G prod. - G react. = 0 At Equilibrium  G = 0, Q = K For reactions run under “nonstandard conditions”  G =  G o + RTlnQ R = 8.31J/molK T = Temp in K 39

40. Free Energy & Equilibrium At equilibrium, ∆G = 0 and k = Q so you can p lug in 0 for ∆G and k for Q, rearranging the formula you get new formula that lets you quantitatively relate free energy to equilibrium. Like wise, you can relate the pressure of reactants and products to the equilibrium constant k with the following equation: 40

41. Quantitative Relationship Between Free Energy And Equilibrium Constant GoGo K  G o = 0 K = 1  G o < 0 K > 1  G o > 0 K < 1 41

42. Free Energy & Equilibrium Consider the ammonia synthesis reaction Where  G o = -33.3 kJ per mole of N 2 consumed at 25 o C. For each of the following mixtures of reactants and products at 25 o C, predict the direction in which the system will shift to reach equilibrium. 1.P NH3 = 1.00 atm, P N2 = 1.47 atm, P H2 = 1.00x10 -2 atm 2.P NH3 =1.00 atm, P N2 =1.00 atm, P H2 = 1.00 atm 42

43. Temperature Dependence Of K  G o = -RTln(K) =  H o - T  S o Rearrangement gives A plot of ln(K) vs 1/T yields a straight line (Y=mx+b) 43

44. Free energy And Work Free energy is energy free to do work The maximum amount of work possible at a given temperature and pressure is: w max =  G Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work. 44

45. Summary of Thermodynamic Laws First Law: You can’t win, you can only break even Second Law: You can’t break even 45