1. Chapter 17 1

2. The Austrian physicist Ludwig Boltzmann introduced a model to relate the total number of microstates (the multiplicity, W) to entropy (S). S = k ln W Boltzmann constant (k) units on S is J/K k = 1.38 x 10  23 J/K Gas constant Avogadro constant Microstates and Entropy Boltzmann was the founding father of statistical mechanics, a completely new way of thinking about theoretical physics. His work was ridiculed by his fellow professors.

3. 3 Standard Entropy Difficult to count the number of microstates directly. Measured by calorimetery. Standard entropy (S  ) absolute entropy of a substance at 1 atm (typically at 25  C) Units of J K -1 mol -1 All positive values. Absolute values (unlike  H which is relative)

4. 4 Why is there a –S? S = k ln W Al + 6H 2 O Al(H 2 O) 6 (aq) 3+ 2+

5. Chapter 17 5

6. Spontaneity Spontaneous: process that does occur under a specific set of conditions. Nonspontaneous: process that does not occur under a specific set of conditions. The reverse of a spontaneous process is a nonspontaneous one. Spontaneity depends on temperature. 6

7. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. Predicting Spontaneity  H rxn = H products – H reactants H products < H reactants  H < 0 H products > H reactants  H > 0 ExothermicEndothermic Enthalpically favorable Enthalpically unfavorable If we mix reactants and products together will the reaction occur? 7 Not enough information. We need to know the change in entropy to predict if a reaction will spontaneously occur.

8. Entropy (S) Is a measure of the number of specific ways (microstates) in which a thermodynamic system may be arranged. Predicting Spontaneity  S rxn = S products – S reactants S products < S reactants  S < 0 S products > S reactants  S > 0 Entropy decreasesEntropy increases Entropically unfavorable Entropically favorable If we mix reactants and products together will the reaction occur? 8 Still not enough information. We need to know total entropy change to predict if a reaction will spontaneously occur.

9. If we mix reactants and products together will the reaction occur? Predicting Spontaneity Nonspontaneous process:  S univ =  S sys +  S surr Equilibrium :  S univ = 0  S univ < 0 Spontaneous process:  S univ > 0  S univ =  S sys +  S surr Can calculate from: S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - If we know  S surr we can calculate  S univ 9

10.  S univ =  S sys +  S surr Can calculate from: S0S0 rxn nS 0 (products) =  mS 0 (reactants)  - If we know  S surr we can calculate  S univ Predicting Spontaneity  S 0 rxn surrounding system  S surr Surrounding = everything in the universe except the system. Very, very difficult to measure! If not impossible. 10

11. Exothermic Reaction Endothermic Process Heat and Entropy  S surr > 0  H sys Surroundings + heat = ↑ S 0 <  H sys  S surr < 0  H sys Surroundings - heat = ↓ S 0 >  H sys  S surr ∝ -  H sys 11

12. Heat and Entropy Heat released by the system increases the disorder of the surroundings.  S surr ∝ -  H sys The effect of -  H sys on the surroundings depends on temperature: – At high temperature, where there is already considerable disorder, the effect is muted – At low temperature the effect is much more significant – The difference between tossing a rock into a calm pool (low T) and a storm-tossed ocean (high T)  S surr = -  H sys T 12

13.  S univ =  S sys +  S surr Predicting Spontaneity  S surr = -  H sys T  S univ =  S sys + -  H sys T -T  S univ = -T  S sys +  H sys Substitution: Multiply by -T: -T  S univ =  H sys - T  S sys Rearrange: Both in terms of the system. This equation relates  S univ to  H sys and  S sys. 13

14. Gibbs Free Energy -T  S univ =  H sys - T  S sys Josiah Willard Gibbs (1839-1903)  G =  H sys - T  S sys First American Ph.D. in Engineering (Yale, 1863) Praised by Albert Einstein as "the greatest mind in American history" Died at 64 from an acute intestinal obstruction. Gibbs free energy (  G)- AKA Free energy. Relates S, H and T of a system. Can be used to predict spontaneity. G is a state function. in K 14

15. Gibbs Free Energy  G =  H sys - T  S sys Gibbs free energy (  G)- Can be used to predict spontaneity. For a constant temperature and constant pressure process:  G < 0 The reaction is spontaneous in the forward direction.  G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.  G = 0 The reaction is at equilibrium.  G = -T(+  S univ )  G = -T(-  S univ )  S univ > 0 -T  S univ =  H sys - T  S sys  S univ < 0  G = -T(  S univ ) = 0  S univ = 0 15

16. Gibbs Free Energy  G =  H - T  S If you know  G for reactants and products then you can calculate if a reaction is spontaneous. If you know  G for two reaction then you can calculate if the sum is spontaneous. If you know  S,  H and T then you can calculate spontaneity. Can predict the temperature when a reaction becomes spontaneous. If you have  H vap or  H fus and  S you can predict boiling and freezing points. If you have  H vap or  H fus and T you can predict the entropy change during a phase change. Can predict equilibrium shifts. 16

17. Standard Free Energy Changes The standard free-energy of reaction (  G 0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn aA + bB cC + dD G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - G0G0 rxn d  G 0 (D) f c  G 0 (C) f = [+] - b  G 0 (B) f a  G 0 (A) f [+] Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f  G 0 of any element in its stable form is zero. f 17

18. aA + bB cC + dD Standard Free Energy Changes  G is a state function so free energy can be calculated from the table of standard values just as enthalpy and entropy changes. G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - Standard free energy of formation (  G 0 ) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f 18

19. Standard Free Energy Changes From appendix 3: KClO 3 (s)  G  f = -289.9 kJ/mol KCl(s)  G  f = -408.3 kJ/mol O 2 (g)  G  f = 0 kJ/mol 2KClO 3 (s)  2KCl(s) + 3O 2 (g) Calculate the standard free-energy change for the following reaction: G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  -  G 0 rxn = [2(  408.3 kJ/mol) + 3(0)]  [2(  289.9 kJ/mol)]  G 0 rxn =  816.6  (  579.8)  G 0 rxn =  236.8 kJ/mol  G 0 rxn < 0 Yes! Is the reaction spontaneous? 19

20. Example 20 17.4 Calculate the standard free-energy changes for the following reactions at 25°C. (a) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) From appendix 3: CH 4 (g)  G  f = -50.8 kJ/mol O 2 (g)  G  f = 0 kJ/mol CO 2 (g)  G  f = -394.4 kJ/mol H 2 O (l)  G  f = -237.2 kJ/mol ΔG° rxn = [ΔG° f (CO 2 ) + 2ΔG° f (H 2 O)] - [ΔG° f (CH 4 ) + 2ΔG° f (O 2 )] ΔG° rxn =[(-394.4 kJ/mol) + (2)(-237.2 kJ/mol)] - [(-50.8 kJ/mol) + (2) (0 kJ/mol)] ΔG° rxn = -818.0 kJ/mol Spontaneous? Yes.

21. Example 21 17.4 Calculate the standard free-energy changes for the following reactions at 25°C. (b) 2MgO(s) 2Mg(s) + O 2 (g) From appendix 3: MgO(s)  G  f = -569.6 kJ/mol O 2 (g)  G  f = 0 kJ/mol Mg(s)  G  f = 0 kJ/mol ΔG° rxn = [2ΔG° f (Mg) + ΔG° f (O 2 )] - [2ΔG° f (MgO)] ΔG° rxn = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-569.6 kJ/mol)] ΔG° rxn = 1139 kJ/mol Spontaneous? No. But the reverse is… 2Mg(s) + O 2 (g) 2MgO(s)

22. More  G° Calculations G0G0 rxn n  G 0 (products) f =  m  G 0 (reactants) f  - To predict spontaneity of any rxn: 1)Pick any reactants and products. 2)Write a balanced equation. 3)Calculate  G 0 rxn. 4)Is the reaction spontaneous? Appendix 3 22

23. Another Example + O 2 C(s, diamond) + O 2 (g) CO 2 (g)  G° rxn = -396.4 kJ Therefore, diamonds are contributing to global warming! very slowly ΔG° rxn = [ΔG° f (CO 2 )] - [ΔG° f (C, diamond) + ΔG° f (O 2 )] From appendix 3: C, diamond(s)  G  f = 2.9 kJ/mol O 2 (g)  G  f = 0 kJ/mol CO 2 (g)  G  f = -394.4 kJ/mol 23

24. 24 More  G° Calculations Similar to  H°, one can use the  G° for various reactions to determine  G° for the reaction of interest (a “Hess’ Law” for  G°) Hess’ Law- states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. What is the  G° for this reaction: Given: C(s, diamond) + O 2 (g) CO 2 (g)  G° = -397 kJ C(s, graphite) + O 2 (g) CO 2 (g)  G° = -394 kJ

25. More  G° Calculations What is the  G° for this reaction: C(s, graphite) + O 2 (g) CO 2 (g)  G° = -394 kJ CO 2 (g) C(s, graphite) + O 2 (g)  G° = +394 kJ Given: C(s, diamond) + O 2 (g) CO 2 (g)  G° = -397 kJ CO 2 (g) C(s, graphite) + O 2 (g)  G° = +394 kJ C(s, diamond) C(s, graphite)  G° = -3 kJ  G° rxn < 0…..rxn is spontaneous 25

26. Alternative  G Calculation Is the following reaction spontaneous at 298 K? (assume standard conditions)  H° f (kJ/mol) S° (J/mol.K) KClO 3 (s)-397.7143.1 KClO 4 (s)-432.8151.0 KCl (s)-436.782.6 Given:  G° rxn =  H° rxn - T  S° rxn First Calculate Given Then Find 26

27. Alternative  G Calculation  H° f (kJ/mol) S° (J/mol.K) KClO 3 (s)-397.7143.1 KClO 4 (s)-432.8151.0 KCl (s)-436.782.6 27

28. Alternative  G Calculation  H° f (kJ/mol) S° (J/mol.K) KClO 3 (s)-397.7143.1 KClO 4 (s)-432.8151.0 KCl (s)-436.782.6  H° rxn = -144 kJ  S° rxn = -36.8 J/K  G° rxn < 0…..rxn is spontaneous at 298 K Enthalpically favorable Entropically unfavorable What about at 5000 K?  G° rxn = 50 kJ  G° rxn > 0…rxn is nonspontaneous at 5000 K (Assuming the same  H and S) 28

29.  G Temperature Dependence  G° rxn = 50 kJ at 5000 K  G° rxn = -133 kJ at 298 K Spontaneous Nonspontaneous Reaction spontaneity is a temperature dependent phenomenon!  G rxn =  H rxn - T  S rxn 29

30.  G Temperature Dependence  G =  H - T  S Enthalpy:  H rxn < 0 The reaction is enthalpically favorable. Entropy:  S rxn > 0 The reaction is entropically favorable. Need both to predict spontaneity. And sometimes temperature!  G < 0Spontaneous  G > 0Nonspontaneous  G = 0Equilibrium 1) If  H 0, then  G is negative at all T 3) If  H < 0 and  S < 0, then  G depends on T 2) If  H > 0 and  S > 0, then  G depends on T 4) If  H > 0 and  S < 0, then  G is positive at all T 30

31.  G =  H - T  S  G < 0Spontaneous  G > 0Nonspontaneous  G = 0Equilibrium 3)  H < 0 and  S < 0 If  H < T  S, then  G is positive. If  H > T  S, then  G is negative.  G Temperature Dependence 2)  H > 0 and  S > 0 If  H < T  S, then  G is negative. If  H > T  S, then  G is positive. Nonspontaneous at high T Spontaneous at low T (Enthalpically favorable, entropically unfavorable) (Enthalpically unfavorable, entropically favorable) Spontaneous at high T Nonspontaneous at low T 31

32. 32  G =  H - T  S  G Temperature Dependence

33. 33  G Temperature Dependence Nonspontaneous Spontaneous Four possible scenarios: (1)  H 0 (2)  H > 0,  S > 0 (3)  H < 0,  S < 0 (4)  H > 0,  S < 0

34. exothermic system becomes more disordered exothermic system becomes more ordered endothermic system becomes more disordered  S univ and Spontaneous Reactions  G =  H - T  S Class 1: Class 3: Class 2: 34

35. 35 Predicting T from Gibbs Equation  G° rxn = 50 kJ at 5000 K  G° rxn = -133 kJ at 298 K Spontaneous Nonspontaneous Reaction spontaneity is a temperature dependent phenomenon! At what temperature will the reaction become spontaneous? Find T where  G changes from positive to negative. I.e. when  G =0.  G =  H – T  S = 0 T =  H/  S

36. 36 Predicting T from Gibbs Equation At what T is the following reaction spontaneous? Br 2 (l) Br 2 (g) Given:  H°= 30.91 kJ/mol  S°= 93.2 J/mol.K  G =  H – T  S = 0 T =  H/  S T = (30.91 kJ/mol) /(93.2 J/mol.K) T = 331.7 K  H > 0,  S > 0 The reaction will be spontaneous when T > 331.7 K

37. Predicting T from Gibbs Equation At what T is the following reaction spontaneous? CaCO 3 (s) CaO (s) + CO 2 (g)  G 0 =  H 0 – T  S 0 = 0 T =  0 H/  S 0  H 0 = 177.8 kJ/mol  S 0 = 160.5 J/K·mol  G 0 = 0 at 835 o C Equilibrium Pressure of CO 2 37

38. Chapter 11 38

39. Liquid ↔ Gas At time = 0At time > 0At time = ∞ Liquid # of molecules in = # of molecules out At equilibrium! rate of evaporation = rate of condensation  G =  H - T  S  G < 0Spontaneous  G > 0Nonspontaneous  G = 0Equilibrium 39

40. 40 Gibbs Equation and Phase Change Molar heat of vaporization (  H vap ) is the energy required to vaporize 1 mole of a liquid at its boiling point. H 2 O (l) H 2 O (g) GG =  H – T  S = 0 If we know  H vap and boiling point we can calculate  S!  S = T HH = 40.79 kJ/mol 373 K SS  H vap = 40.79 kJ/mol BP(H 2 O) = 373 K

41. Example 41 17.5 The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid → liquid and liquid → vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. solid liquid vapor

42. Example 42 17.5 The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid → liquid and liquid → vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. Molar heat of vaporization (  H vap ) is the energy required to vaporize 1 mole of liquid to gas. Molar heat of fusion (  H fus ) is the energy required to melt 1 mole of solid. solid → liquid → vapor  H vap  H fus

43. Example 43 17.5 The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid → liquid and liquid → vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. solid → liquid → vapor  H vap  H fus GG =  H – T  S = 0  S = T HH

45. Chapter 17 45