1 Entropy & Gibbs Free Energy Chapter 19

2 The heat tax No matter what the process, heat always lost to surroundings No matter what the process, heat always lost to surroundings More energy required to recharge batteries than energy available for work More energy required to recharge batteries than energy available for work

3 Spontaneity Spontaneous process does NOT require outside energy Spontaneous process does NOT require outside energy –Such as a car rusting Non-spontaneous process requires external energy Non-spontaneous process requires external energy –Imagine trying to “un-rust” the jalopy above Not all spontaneous processes exothermic Not all spontaneous processes exothermic –Ice melting above melting point

4 Entropy Disorder/randomness in a system Disorder/randomness in a system Entropy increases in universe (2 nd Law) Entropy increases in universe (2 nd Law) –We’re all going to dissolution

5 Entropy: the “book” definition Entropy (S) = a thermodynamic function that increases with number of energetically equivalent ways to arrange components of a system to achieve a particular state Entropy (S) = a thermodynamic function that increases with number of energetically equivalent ways to arrange components of a system to achieve a particular state HUH?! HUH?!

6 Towards greater entropy Systems process in direction that has largest energetically equivalent ways to arrange its parts Systems process in direction that has largest energetically equivalent ways to arrange its parts Which configuration has greater entropy; i.e., largest energetically equivalent ways to arrange itself? Which configuration has greater entropy; i.e., largest energetically equivalent ways to arrange itself?

7 Ludwig Boltzmann Distribution of energy over different energy states used as a way to calculate entropy Distribution of energy over different energy states used as a way to calculate entropy S = k  log W k = R/N = 1.38 x J/K = Boltzmann constant (R = J/mol●K & N = Avogadro’s # = x atoms/mole) k = R/N = 1.38 x J/K = Boltzmann constant (R = J/mol●K & N = Avogadro’s # = x atoms/mole) W = # of energetically equivalent ways to arrange components of system (unitless) W = # of energetically equivalent ways to arrange components of system (unitless)

8

9 Standard molar entropy, S ° For standard state elements (metals, gases, etc.) molar enthalpy of formation, H° f = 0 J/mol For standard state elements (metals, gases, etc.) molar enthalpy of formation, H° f = 0 J/mol For entropy, it’s different: For entropy, it’s different: –3 rd law of thermodynamics: The entropy of a perfect crystal at absolute zero (0K) is zero. The entropy of a perfect crystal at absolute zero (0K) is zero. Since only one way to arrange (W = 1) Since only one way to arrange (W = 1)  S = k  log W = k  log 1 = 0  S = k  log W = k  log 1 = 0

10 More on entropy Gas entropies are much larger than those for liquids Gas entropies are much larger than those for liquids Liquid entropies are larger than solids Liquid entropies are larger than solids For ex: For ex: –I 2(s) vs. Br 2(l) vs. Cl 2(g) standard entropies: 116.1, 152.2, J/mol●K, respectively –Solid C vs. gaseous C: 5.6 vs J/mol●K

11 Even more on entropy! Generally, larger molecules have larger entropy than smaller molecules Generally, larger molecules have larger entropy than smaller molecules –Due to greater molar mass Molecules with more complex structures have larger entropies than simpler molecules Molecules with more complex structures have larger entropies than simpler molecules –Since there are more ways for molecule to twist, rotate, vibrate in space For ex: For ex: –CH 4, C 2 H 6, C 3 H 8 : 186.3, 229.2, J/mol  K –Ar, CO 2, C 3 H 8 : 154.9, 213.7, J/mol  K

12  S ° rxn Calculating standard changes in entropy Calculating standard changes in entropy Similar to calculating  H ° rxn Similar to calculating  H ° rxn –Don’t forget molar coefficients!  S ° rxn = S ° products - S ° reactants  S ° rxn = S ° products - S ° reactants All reactants & products in standard states All reactants & products in standard states –J/(mol  K)

13 Appendix L, pages A Thermodynamic values: ∆H, ∆S, & ∆G Thermodynamic values: ∆H, ∆S, & ∆G

14 Problem Compute  S ° rxn for the following equation: Compute  S ° rxn for the following equation: 4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) 4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) Given: Given: (J/mol  K) = ammonia (J/mol  K) = ammonia = oxygen gas = oxygen gas = nitrogen monoxide = nitrogen monoxide = water = water

15

16 Entropy changes and spontaneity According to the 2 nd Law of Thermodynamics: According to the 2 nd Law of Thermodynamics: –“A spontaneous process is one that results in an increase of entropy in the universe.”  S° univ =  S° system +  S° surrounding For a spontaneous process: For a spontaneous process: –  S° univ > 0 For a system at equilibrium: For a system at equilibrium: –  S° univ = 0 For a nonspontaneous process: For a nonspontaneous process: –  S° univ < 0

17 Two questions Work on these individually and turn them in: Work on these individually and turn them in: –Does the entropy of surrounding increase or decrease in an exothermic process? –Does the entropy of surrounding increase or decrease in an endothermic process?

18 Temperature dependence of  S surr  S univ =  S sys +  S surr Therefore, for water freezing: Therefore, for water freezing: –  S sys is negative (exothermic) –  S surr is positive (endothermic)  S is large at low temperatures (closer to freezing point): has very little E, so big difference made  S is large at low temperatures (closer to freezing point): has very little E, so big difference made  S is small at high temps (further away from fp): already has more E due to higher temp  little difference  S is small at high temps (further away from fp): already has more E due to higher temp  little difference –IOW, magnitude of  S depends on temp (at above 0°C)

19 Quantifying entropy changes in surrounding -  H sys   S surr -  H sys   S surr  S surr  1/T  S surr  1/T  S surr  -  H sys /T  S surr  -  H sys /T  exothermic processes have tendency to be more spontaneous at low temps than high  exothermic processes have tendency to be more spontaneous at low temps than high –They increase entropy of surrounding more if T is small As temp increases (greater E in surrounding),  H decreases, producing a smaller  S surr As temp increases (greater E in surrounding),  H decreases, producing a smaller  S surr

20 Problem C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g) Calculate entropy change in surrounding at 25°C given  H° rxn = kJ Calculate entropy change in surrounding at 25°C given  H° rxn = kJ Determine  S sys, given: Determine  S sys, given: –H 2 O (g) = J/mol  K –CO 2(g) = J/mol  K –C 3 H 8(g) = J/mol  K –O 2(g) = J/mol  K What is  S univ ? What is  S univ ?

21

22 Spontaneous or not? 1Exothermic Less order Spontaneous<0>0Suniv>0 2Exothermic More order Depends on H & S <0<0 More favorable at lower temps 3Endothermic Less order Depends on H & S >0>0 More favorable at higher temps 4Endothermic More order Nonspontaneous >0<0Suniver<0

23 Examples of each type Type I Type I –Exothermic (  H sys 0) Combustion reactions Combustion reactions C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g) Type IV Type IV –Endothermic (  H sys >0 ) and more order (  S sys 0 ) and more order (  S sys <0) N 2(g) + 2H 2(g)  N 2 H 4(l)

24 Examples of each type (cont.) Type II Type II –Exothermic (  H sys <0 ) and more order (  S sys <0) Temperature dependent Temperature dependent –More favorable at lower temperatures N 2(g) + 3H 2(g)  2NH 3(g)

25 Examples of each type (cont.) Type III Type III –Endothermic (  H sys >0 ) and less order (  S sys >0) Temperature dependent Temperature dependent –More favorable at higher temperatures NH 4 Cl (s)  NH 3(g) + HCl (g)

26 Gibbs Free Energy Measures spontaneity of a process with evaluation of only the system Measures spontaneity of a process with evaluation of only the system –No longer needs evaluation of surrounding “Free energy” =“available energy” to do work “Free energy” =“available energy” to do work –Sum of energies available from dispersal of energy and matter

27 G  G (  G 0)  G (  G 0)  G (  G>0) = non-spontaneous process (  S 0) = non-spontaneous process (  S<0)

28 The effect of  H,  S, & T on ∆G Type I Type I –Exothermic (-  H) –Entropy > 0 (+  S) –-  G Spontaneous at all temps Spontaneous at all temps 2N 2 O (g)  2N 2(g) + O 2(g)

29 The effect of  H,  S, & T on ∆G Type IV Type IV –Endothermic (+  H) –Entropy < 0 (-  S) –+  G Nonspontaneous at all temps Nonspontaneous at all temps 3O 2(g)  2O 3(g)

30 The effect of  H,  S, & T on ∆G Type II Type II –Exothermic (-  H) –Entropy < 0 (-  S) –  G is temp-dependent H 2 O (l)  H 2 O (s) -  H Water freezing is temp-dependent  low temps high temps

31 The effect of  H,  S, & T on ∆G Type III Type III –Endothermic (+  H) –Spontaneous (+  S) –  G is temp-dependent H 2 O (l)  H 2 O (g) +  H Water boiling is temp-dependent  high temps low temps

32 Calculating  G ° rxn Calculate  G ° rxn (at 25°C) for Calculate  G ° rxn (at 25°C) for SO 2(g) + ½ O 2(g)  SO 3(g) SO 2 ;  H ° f = kJ/mol &  S ° = J/(mol  K) SO 2 ;  H ° f = kJ/mol &  S ° = J/(mol  K) O 2 ;  H ° f = ?,  S ° = 205.2J/(mol  K) O 2 ;  H ° f = ?,  S ° = 205.2J/(mol  K) SO 3 ;  H ° f = kJ/mol,  S ° = J/(mol  K) SO 3 ;  H ° f = kJ/mol,  S ° = J/(mol  K) Is it spontaneous under standard conditions? Is it spontaneous under standard conditions?

33

34 Calculating  G ° rxn from free energies of formation Free E of formation of pure elements in standard states = 0 kJ/mol Free E of formation of pure elements in standard states = 0 kJ/mol  G ° rxn =  G ° products -  G ° reactants  G ° rxn =  G ° products -  G ° reactants CH 4(g) + 8O 2(g)  CO 2(g) + 2H 2 O (g) CH 4(g) + 8O 2(g)  CO 2(g) + 2H 2 O (g) ? ? Calculate  G ° rxn Calculate  G ° rxn

35

36 Determine  G ° rxn for 3C (s) + 4H 2(g)  C 3 H 8(g) Given: Given: C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g)  G ° rxn = kJ C (s) + O 2(g)  CO 2(g)  G ° rxn = kJ 2H 2(g) + O 2(g)  2H 2 O (g)  G ° rxn = kJ

37

38 Non-standard conditions  G ° rxn is at standard conditions  G ° rxn is at standard conditions –25 °C, 273 K, 1 atm, 1M, etc.  G rxn = non-standard conditions  G rxn = non-standard conditions –Explains why water evaporates off floor even though  G ° rxn = 8.59 kJ/mol &  is a non- spont process. Partial pressure of water well below 1 atm  Non-standard condition! Partial pressure of water well below 1 atm  Non-standard condition!

39 Free E change under non-standard conditions  G rxn =  G ° rxn + RT(lnQ)  G rxn =  G ° rxn + RT(lnQ) Q = rxn quotient, R = J/(mol  K) Q = rxn quotient, R = J/(mol  K)

40 Problem Compute  G rxn under the following conditions for: Compute  G rxn under the following conditions for: 2NO (g) + O 2(g)  2NO 2(g) –P NO = atm; P O 2 = atm; P NO 2 = 2.00 atm –  G ° rxn = kJ –At 25°C

41

42 Relating  G ° rxn to K  G rxn =  G ° rxn + RT(lnQ) At equilibrium, Q = K &  G rxn = 0 0 =  G ° rxn + RT(lnK)  G ° rxn = -RT(lnK) When K<1 (reactant-favored) When K<1 (reactant-favored) –lnK = “-” &  G ° rxn = “+”  spontaneous in reverse direction  spontaneous in reverse direction When K>1 (product-favored) When K>1 (product-favored) –lnK = “+” &  G ° rxn = “-’  spontaneous in fwd direction  spontaneous in fwd direction When K = 1, lnK = 0 &  G ° rxn = 0 When K = 1, lnK = 0 &  G ° rxn = 0 –  equilibrium under standard conditions

43 Temp dependence of K How does one obtain the equation on the left? How does one obtain the equation on the left? Useful for obtaining  H ° rxn &  S ° rxn Useful for obtaining  H ° rxn &  S ° rxn Does the equation look familiar? Does the equation look familiar? How would we plot this? How would we plot this?