1. Chapter 19 Thermodynamics Ch 14 Kinetics: How fast do reactions go? Ch15 Equilibrium: How far do reactions proceed? Ch 19 Chemical Thermodynamics: What makes reactions go?

2. Thermodynamics The study of the relationships between heat, work, and the energy of a system. You may recall that changes can be exothermic (-  H) or endothermic (+  H)

3. Exothermic: activation energy (A) is absorbed to break bonds, and energy is released as new bonds form (B). Heat is released to the surroundings, since B>A Endothermic: activation energy (A) is absorbed to break bonds, and energy is released as new bonds form (B). Heat is absorbed from the surroundings, since B<A -H-H +H+H

4. Exothermic can usually continue once begun since they can recycle some of the energy released (B) as activation energy (A) to keep the reaction going. Changes like this are said to be spontaneous. Endothermic do not release enough energy (B) to provide all the activation energy to keep reaction going, so outside energy must usually be added to drive the reaction. Changes like this are usually not spontaneous. ?

5. Actually the AP folks prefer you to use the work favorable in replace of spontaneous. Either way it means a reaction that will continue on its own once begun. So exothermic reactions are favorable since they provide their own activation energy. ? To clarify: Spontaneous in the English language implies “a process which begins on its own – without cause”. Spontaneous in chemical vernacular means a process that continues on its own once it is caused to begin – by an outside force. In order to avoid eliminate confusion the AP folks avoid using the word “spontaneous” even though the term is still commonly used in chemistry. Instead you will see the term “favorable”.

6. Exothermic: combustion Spontaneous once begun Endothermic: Cooking, Not spontaneous, cooking stops once heat is removed. favorable Not favorable

7. But, some endothermic reactions drive themselves! How? Dissolving proceeds on its own without outside intervention. Its Entropy in action favorable Entropy is “S”

9. Crash course chemistry #17: Energy & Chemistry Crash course chemistry #18: Enthalpy Crash course chemistry #20: Entropy Review of chapter 5 topics Chapter 19 topic: Gibbs Free energy

10. 19.1 Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously move to just one vessel. (favorable)

11. Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. Spontaneous processes are irreversible. favorable

12. Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C it is spontaneous for ice to melt. Below 0  C the reverse process is spontaneous. favorable

14. 19.2/3 Entropy Entropy can be thought of as a measure of the randomness or disorder of a system. Entropy increases as the number of modes of motion in molecules (microstates) increases Vibrational Rotational Translational

15. The Entropy of a pure crystalline substance at absolute zero is zero. why? It has only one possible microstate. As temperature increases the molecules begin to move and there become multiple possible microstates.

16. The Second law (of thermodynamics): Entropy in the universe always increases Spontaneous processes happen when randomness in the universe increases The first law was “Conservation of Energy” (  E system =  E surroundings ): Chapter 5 Because rusting is spontaneous, overall  S must be positive. therefore the entropy of the surroundings must increase more than the entropy decrease of the system.

17. Molecules have more entropy (disorder) when: 1)Phase Changes from: S  L  G Example: Sublimation CO 2 (s)  CO 2 (g) 2)Dissolving occurs (solution forms): Example: NaCl(s)  Na + (aq) + Cl - (aq) Matter spreads out (more microstates) Matter spreads out (more microstates)

18. 3) Temperature increases Example: Fe(s) at 0 o C  Fe(s) at 25 o C 4) For Gases ONLY, when Volume increases or Pressure decreases Examples: 2 Liters He(g)  4 Liters He(g) 3 atm He(g)  1 atm He(g) Molecules have more entropy (disorder) when: Energy spreads out (more microstates) matter spreads out (more microstates)

19. 5) Rx results in more molecules/moles of gas Examples: 2NH 3 (g)  N 2 (g) + 3H 2 (g) CaCO 3 (s)  CaO(s) + CO 2 (g) N 2 O 4 (g)  2 NO 2 (g) This one is difficult to predict: N 2 (g) + O 2 (g)  2 NO(g) Molecules have more entropy (disorder) when: matter spreads out (more microstates)

20. 6)When there are more moles Example: 1 mole H 2 O(g)  2 moles H 2 O(g) 7) When there are more atoms per molecule Examples: 1 mole Ar(g) vs 1 mole HCl(g) 1 mole NO 2 (g) vs 1 mole N 2 O 4 (g) Molecules have more entropy (disorder) when: More matter is more spread out (more microstates)

21. 8) When an atom has a bigger atomic number 1 mole He(g) vs. 1 mole Ne (g) Molecules have more entropy (disorder) when: More matter (P,N,e) = more spread out (more microstates) In spite of all of these, entropy changes during phase changes is the most important. (solid  liquid or Aqueous  gases)

22. Again notice the typical change in entropy reduces to changes of state

25. Go figure: Why does the plot show vertical jumps at the melting and boiling points? Separation of molecules produce the greatest increase in entropy (motions and microstates)

26. 19.4 Entropy changes / Standard Entropies Standard Conditions: 298 K, 1 atm, 1 Molar The values for Standard Entropies (S o ) are expressed in J/mol-K. Note: Increase with increasing molar mass.

27. Entropy Changes  S o =  n S o products -  m  S o reactants Be careful: S°units are in J/mol-K Note for pure elements: Even though: (Not kJ’s) (recall that we ignored elements in Hess’ law probs?) To clarify:  S has an actual value since its based on number of possible microstates, while  H is a relative value based on loss or gain of energy (not energy actually stored in the molecule) so the  H for the lowest energy state element is set at zero.

28. Appendix C values Al 2 O 3(s) = 51.00 Jmol - K - H 2(g) = 130.58 Al (s) = 28.32 H 2 O (g) = 88.83

29. Reference Tables Heat exchanged during temp change (calorimetry) Entropy change Enthalpy change (Hess’ law) Free energy (Gibbs) change Up next! Free energy and the equilibrium constant (coming soon)

30. 19.5 Gibbs Free Energy Enthalpy Enthalpy and entropy both contribute to help drive reactions Exothermic changes lose heat to the surroundings Matter tends to spread out increasing entropy

31. 19.5 Gibbs Free Energy Enthalpy Reactions which release heat are favorable Entropy tends to increases as disorder increases Entropy increases as temperature increases T is combined with S into entropy term T  S

32. 19.5 Gibbs Free Energy Enthalpy The two are Combined into Gibb’s “Free” energy (energy free to do work on the surroundings)

33. 19.5 Gibbs Free Energy Enthalpy  H + - (T  S) Taking the negative of the entropy term allows us to combine the two terms into one  H - T  S  G =  H - T  S

34. 19.5 Gibbs Free Energy Enthalpy You should note that Free energy released into the surroundings (-  G) translates into an increase in the entropy of the universe as a whole (+  S univ ).

35. 19.5 Gibbs Free Energy Enthalpy Spontaneous changes occur when the universe becomes more disorderly.

36. 19.5 Gibbs Free Energy Enthalpy Since the universe is a difficult thing to view as a whole, we can focus just on the changes occurring in the system:  G

37. 19.5 Gibbs Free Energy Use  G to decide if a process is spontaneous  G = negative value = spontaneous  G o =  H o – T  S o favorable Favorable – free energy released to the surroundings?

38. 19.5 Gibbs Free Energy  G = zero = at equilibrium  G o =  H o – T  S o For reversible reactions

39. 19.5 Gibbs Free Energy  G = positive value = not spontaneous Note: o under standard conditions Equation can be used without also.  G o =  H o – T  S o unfavorable

40. Gibbs Free Energy 1.If  G is negative  G = is the maximum amount of energy ‘free’ to do work by the reaction 2. If  G is positive  G = is the minimum amount of work needed to make the reaction happen  G o =  H o – T  S o

41. Gibbs Free Energy In our tables, units are:  G o = kJ/mol  H o = kJ/mol  S o = J/mol-K  G o =  H o – T  S o

42. What causes a reaction to be spontaneous? Think Humpty Dumpty System tend to seek:  Minimum Enthalpy Exothermic Rx,  H = negative  Maximum Entropy More disorder,  S = positive Because:  G o =  H o – T  S o - = (-) - (+) favorable Energy spreads out (into the surroundings) matter spreads out

44. Notice the unit conversion

48. Subtracting a positive entropy value makes G a larger negative

49. 19.6 Free Energy and Temperature By knowing the sign (+ or -) of  S and  H, we can get the sign of  G and determine if a reaction is spontaneous.

50.  G o =  H o – T  S o =  –  – (+) Ex1: Combustion - Exothermic with production of gases CH (s) compound + O 2(g)  CO 2(g) + H 2 O (g)  H o / +  S o  G o always negative Favorable under all conditions Continues once begun. When both enthalpy and entropy changes are favorable, the overall reaction must be favorable under all conditions!

51. Even though reactions may be thermodynamically favorable, they may still not proceed if their reaction rates are very slow. Reactions with high activation energies may be too slow and still require a catalyst to speed things up! These reactions are said to be under “kinetic control” Recall from chapter 14 that speed of a reaction depends on activation energy?

53.  G o =  H o – T  S o =  +  – (–) Ex2: Photosynthesis - endothermic with production of a solid CO 2(g) + H 2 O (g)  CH (s) compound + O 2(g)  H o / -  S o  G o always positive “unfavorable under all conditions Stops without input of light When enthalpy and entropy changes are both unfavorable the reaction is unfavorable not matter what the conditions. Continuous energy input is necessary to drive the reaction.

54. How can unfavorable (non-spontaneous) reactions occur? Drive them with an outside energy source: EX: Electrolytic decomposition of water 2H 2 O + energy  2H 2 + O 2 Batteries can be recharged using outside energy

55. Coupling reactions Another way to make unfavorable reaction become favorable is to couple reactions. Hess’s law tells us that the overall  G is the sum of the individual values Ex: CaCO 3  CaO + CO 2  G = +130 kJ But by heating with carbon: C + O 2  CO 2  G = -394 CaCO 3 + C + O 2  CaO + 2CO 2  G = -264 kJ … and the reaction proceeds favorably Calcium oxide is used to make concrete and mortar for building

57.  G o =  H o – T  S o =  –  – (–) Ex3: Haber process - exothermic with decrease of molecule # N 2(g) + 3H 2 (g)  2NH 3(g)  H o / -  S o at lower temps  –  H o ) > (– T  S o ) so  G o is (-) (larger -) – (smaller -)  (-) + (+) = (-) Reaction will proceed spontaneously to the right Products > reactants at equilibrium  G o =  H o – T  S o A “mixed” problem: one favorable factor, the other unfavorable. The reaction is exothermic (favorable energy release) but we are combining to form fewer molecules (entropy is decreasing – unfavorable) At lower temps the unfavorable energy term is minimized

58.  G o =  H o – T  S o =  –  – (–) Ex: Haber process - exothermic with decrease of molecule # N 2(g) + 3H 2 (g)  2NH 3(g)  H o / -  S o at higher temps  –  H o ) < (– T  S o ) so  G o is (+) (smaller -) – (larger -)  (-) + (+) = (+) Reaction will not proceed spontaneously Or equilibrium shifts so that reactants > products at equilibrium  G o =  H o – T  S o At higher temps the unfavorable entropy term dominates.

59. H Enthalpy S Entropy I think of the enthalpy/entropy interaction as scales. In a mixed problem you need to figure out which has a greater influence. Recognize that only factor that can be changed though is S, by increasing or decreasing temperature.

60.  G o =  n G o f products -  m  G o f reactants Standard Free Energy Changes Be careful: Values for  G f  are in kJ/mol  G  can be looked up in tables or calculated from S° and  H .   G o =  H o – T  S o

62.  G o =  H o – T  S o [19.12]  G o = (-  H o ) – (-T  S o ) -T  S becomes larger at higher temps leading to a lower negative  G

64. Q = (P NH3 ) 2 (P N2 )(P H2 ) 3 Q= (1 atm) 2 = 1 (1 atm)(1atm) 3 Kp = 1.07 at 25 o C Q < Kp reaction shifts right

65. Q = (P NH3 ) 2 (P N2 )(P H2 ) 3 Q= (1 atm) 2 = 1 (1 atm)(1atm) 3 Kp =.0038 at 800K Q > Kp reaction shifts left

66. At Equilibrium (Haber process)  G = zero and  H o = T  S o  If  G o = negative then  H o >T  S o If  G o = Positive then  H o <T  S o  G o =  H o – T  S o

67. N 2 + 3H 2   2NH 3 What else does  G  tell us? Since  G  = - 33.3 kJ/mol Rxn at 25 0 C is negative – in a mixture off all three gases with partial pressures of 1.0 atm, the forward reaction is spontaneous, and the reaction shifts right to reach equilibrium.

68. N 2 + 3H 2   2NH 3 And…. Since  G  = +61 kJ/mol Rxn at 500 0 C is positive – in a mixture of all three gases with partial pressures of 1.0 atm, the forward reaction is not spontaneous, and the reaction shifts left to reach equilibrium.

69. Remember: assume that  H and  S at 298K same as at 400K so  G at 400K only depends on change to new temperature T (in T  S)  H o  S o SO 2 = -296.9 kJ/mol248.5 J/molK O 2 = 0 205 J/molK SO 3 = -395.2 kJ/mol256.2 J/molK

70. 19.7 Free Energy and Equilibrium Remember from above: If  G is 0, the system is at equilibrium. So  G must be related to the equilibrium constant, K (chapter 15). The standard free energy,  G°, is directly linked to K eq by:  G o =  RT ln K   Where R = 8.314 J/mol-K

71.  G o =  RT ln K  . Relationships (-) - RT ( ln Large K) If the free energy change is a negative value, the reaction is spontaneous, ln K must be a positive value, and K will be a large number meaning the equilibrium mixture is mainly products -(8.314J/mol - K - )(298K)ln 100 ( - 11.4 kJmol - )

72.  G o =  RT ln K  Relationships (+) - RT ( ln small K) If the free energy change is a positive value, the reaction is not spontaneous, ln K must be a negative value, and K will be a small number meaning the equilibrium mixture is mainly reactants -(8.314J/mol - K - )(298K)ln 0.01 ( + 11.4 kJmol - )

77.  H o  S o CC 4(l) = --139.3 kJ/mol 214.4 J/molK CC 4(g) = -106.7 kJ/mol 309.4 J/molK

79.  H o  S o Br 2(l) = 0 kJ/mol 152.3 J/molK Br 2(g) = +30.71kJ/mol 245.3 J/molK

80.  G o =  RT ln K  kJ =  (8.314 J/molK)(298K) (1kJ/1000J) ln K  ln K  x 10 5  K

81.  G o =  RT ln K