Thermodynamics. Thermodynamically favored reactions ( spontaneous ) Enthalpy Entropy Free energy

Transcription

1 Thermodynamics Thermodynamically favored reactions ( spontaneous ) Enthalpy Entropy Free energy 1

2 Thermodynamically Favored Processes Water flows downhill. Sugar dissolves in coffee. Heat flows from hot object to cold object. A car fender rusts Why not the reverse? 2

3 Thermodynamically Favored Processes Driven by two factors: Enthalpy decrease and/or Entropy increase 3

4 Potential Energy Enthalpy: Many exothermic processes are spontaneous. Lower E a PE reactants time products DH 4

5 Enthalpy CH 4 + 2O 2 CO 2 + H 2 O DH = -890 kj spontaneous (caution rate may be slow) 2H 2 O 2H 2 + O 2 DH = +570 kj Forget about it. 5

6 Enthalpy Recall enthalpy changes are related to bond energies & IM forces of reactants and products. Molecules with strong bonds (high bond energies) are relatively stable and have low potential energy. e.g. H 2 O and CO 2 are often reaction products due to their low PE. 6

7 Enthalpy You might think only exothermic reactions are thermodynamically favored. but H 2 O(s) H 2 O(l) at 25 o C DH = +6.0 kj Ice melts, but is endothermic! 7

8 Entropy (S) Entropy also helps determines whether a reaction is spontaneous. What is it? 8

9 Gedanken Gas Y At 1 atm Gas B At 1 atm What happens when the valve is opened? Why? 9

10 Entropy The gases mix because of Entropy is actually the dispersal of matter and energy. 10

11 Entropy If you drop a glass, what happens? If you drop the broken pieces, do they reform the glass? low S high S 11

12 Changing Entropy: #1 Increasing entropy (+DS) solid liquid gas increased dispersal of matter 12

13 Changing Entropy: #2 Dissolution: H 2 O NaCl(s) Na + (aq) + Cl - (aq) 1.Ionization: increases entropy 2.Mixing: increases entropy 3.Hydration: decreases entropy 13

14 Changing Entropy: #3 Entropy increases in reactions where the number of product molecules is greater than reactant molecules. (matter becomes more dispersed) 4H 3 PO 4 P 4 O H 2 O 14

15 Changing Entropy: #4 Entropy increases greatly in reactions where moles of gas increase. CaCO 3 (s) CaO(s) + CO 2 (g) 0 mol gas 1 mol gas Gases have high entropy due to the dispersal of matter. 15

16 Changing Entropy: #5 Increasing temperature raises the kinetic energy which increases the disorder (entropy). Raising T disperses KE 16

17 Entropy Prediction Is DS positive or negative? CO 2 sublimes Br 2 (l) freezes Sugar is dissolved in water 17

18 Predict DS Is DS positive or negative? 2H 2 (g) + O 2 (g) 2H 2 O(g) 18

19 3 rd Law of Thermodynamics The entropy of a perfect crystalline substance at 0 K is zero. Unlike enthalpy, entropy values are absolute. (no arbitrary reference point) 19

20 Calculating Entropy For the reaction: aa + bb cc + dd Calculate DS from table of standard entropies. (appendix 3; 25 o C and 1 atm) DS o = SnS o (prod) SnS o (react) 20

21 units Calculating Entropy N 2 (g) + 3H 2 (g) 2NH 3 (g) S o (J/mol. K) DS o = 2(193J/K) (1)192J/K 3(131J/K) = -199 J/mol rxn. K Decrease in entropy as expected. 21

22 Calculating Entropy Your turn. H 2 (g) + Cl 2 (g) 2HCl(g) S o (J/mol. K) Does entropy go up or down? 22

23 2 nd Law of Thermodynamics The entropy of the universe increases in a spontaneous reaction and is unchanged in a equilibrium process. 23

24 2 nd Law of Thermodynamics A simple function can be derived from the 2 nd law that describes whether a reaction is spontaneous. G = Gibb s Free Energy 24

25 Gibb s Free Energy DG o = DH o TDS o DG o < 0 reaction spontaneous ( exergonic ) DG o > 0 reaction nonspontaneous (spontaneous in reverse) ( endergonic ) DG o = 0 system at equilibrium 25

26 Standards for DG o Gas 1 atm Liquid pure substance Solid pure substance Elements DG o = 0 Solution 1 M concentration Temp. 25 o C 26

27 Calculating Free Energy 1. From standard DG values fo (appendix 3; 1 mole,1atm, 25 o C). This works just like DH and DS o. fo 2. From DH o and DS o and T using the free energy equation. DG o = DH o TDS o 27

28 Standard DG o H 2 (g) + Br 2 (l) 2HBr(g) DG fo kj/mol most stable form of element DG o = 2(-53.2kJ) 1(0) 1(0) = kj/mol rxn (at 25 o C) Spontaneous!!! 28

29 DG = DH TDS Is a Reaction Spontaneous? DH is + (endo) DH is (exo) DS is + (more random) DS is (less random) 29

30 Spontaneous? Try It! CaCO 3 (s) CaO(s) + CO 2 (g) DH o f (kj/mol) DS o (J/mol. K) Use appendix 3 data (above) to calculate DH o rxn & DS o rxn. Be careful with units. 30

31 Spontaneous? Try It! CaCO 3 (s) CaO(s) + CO 2 (g) 2. Use DH & DS to calc. rxno rxno DG at 25 o C. rxno 3. Is this reaction spontaneous at 25 o C? 4. At what T does this reaction become spontaneous? 31

32 P (CO2) (atm) Free Energy: Interpretation 3 CaCO 3 (s) 2 1 CaO(s) + CO 2 (g) CO 2 hits std. pressure (1atm) 0 Temperature ( o C) 32

33 Coupled Reactions Although some reactions are not thermodynamically favored (+DG), they can be made to happen by coupling to a thermodynamically favored second reaction. 33

34 Coupled Reactions The production of lead from PbS ore by heating is nonspontaneous. Write the net reaction and calculate DG o rxn when this reaction is coupled with the oxidation of sulfur to SO 2. DG rxn o PbS(s) Pb(s) + S(s) +99kJ S(s) + O 2 (g) SO 2 (g) -300kJ 34

35 External Energy External energy can be used to drive reactions where DG is positive. thermal energy electrical energy light energy e.g. photosynthesis DG = 2880 kj/mol rxn hn + 6CO 2 + 6H 2 O C 6 H 12 O 6 + 6O 2 35

36 Free Energy: Reminders 1.Be careful of units (kj vs. J, & T) 2. AP units for chemical reactions: 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) DH o and DG o are kj/mol rxn DS o = J/mol rxn. K 36

37 Free Energy: Reminders 3. DH o & DS o are only exact at 25 o C, but using them to calculate DG o at other temperatures is good enough. (DH o & DS o don t change much with temperature.) 37

38 Free Energy: Reminders 4. Spontaneous reactions (thermodynamically favored) may appear not to occur if the reaction rate is slow due to high activation energy. kinetically controlled reaction 38

39 Controlled by Thermodynamics Controlled by kinetics 39

40 Kinetically Controlled Reaction Graphite is the most stable form of carbon under standard conditions C diamond (s) C graphite (s) DG o = -3 kj/mol rxn (spontaneous) However it is too slow to be observed due to high E a. 40

41 Free Energy: Reminders 5. External energy can be used to drive reactions that are not thermodynamically favored. 2H 2 O(l) 2H 2 (g) + O 2 (g) can be driven by electrical energy Photosynthesis, initiated by photon absorption 41

42 Phase Transitions (e.g. ice water at 0 o C) System is in equilibrium, so: DG o = 0 = DH o -TDS o DS o = DHo T 42

43 Melt Ice at 0 o C DS o = DHo T = = 6010 J 273 K = 22 J/mol. K DH fus T Entropy increases while melting, as expected 43

44 Phase Change: Try It!!! 1.What is the entropy change for freezing benzene at its freezing point of 5.5 o C? (DH fus is 10.9 kj)? 2.Is this process spontaneous? 44

45 Free Energy & Equilibrium aa(aq) + bb(aq) cc(aq) + dd(aq) If reactants start in standard states (1 M), as reaction proceeds, the chemicals are no longer in their standard states. 45

46 Free Energy & Equilibrium As the reaction proceeds, the free energy change (DG) is related to standard free energy change (DG o ) by: DG = DG o + RT lnq reaction quotient, Q, remember? 46

47 Free Energy & Equilibrium aa + bb cc + dd DG = DG o + RT lnq = DG o + RT ln [C]c [D] d [A] a [B] b R = J/mol. K 47

48 Try It N 2 (g) + 3H 2 (g) 2NH 3 (g) P o (atm) What is DG at these initial pressures and 25 o C? (DG o f = -17 kj/mol) Predict the reaction direction. 48

49 Free Energy & Equilibrium At equilibrium: DG = DG o + RT lnq 0 K and DG o = -RT lnk and the net reaction stops. 49

50 Free Energy & Equilibrium DG = DG o + RT lnq zero negative positive Case 1. If reaction is spontaneous (DG o negative), reaction will proceed toward products to a great extent until the RT lnq equals DG o. 50

51 Free Energy & Equilibrium DG = DG o + RT lnq zero positive negative Case 2. If reaction is non- spont. (DG o positive), reaction will proceed toward products to only a small extent until the RT lnq equals DG o. 51

52 Free Energy & Equilibrium At equilibrium: DG o = -RT lnk (K p for gases and K c for solutions) Note: the larger the K, the more negative DG o (reaction goes more toward products). 52

53 Free Energy & Equilibrium DG o = -RT ln K K lnk DG o > < Products favored & spontaneous Reactants favored & nonspontaneous 53

54 Free Energy & Equilibrium DG o = -RT ln K Rearranges to: lnk = - DGo RT and: K = e -DGo /RT RT at room temperature ~ 2.4kJ/mol. So if DG ~ 2.4kJ/mol, K ~ 1, otherwise K will be very large or very small. 54

55 Free Energy & Equilibrium DG o = -RT lnk Measuring DG o (via DH o & DS o ) is a convenient way to measure K for many reactions. 55

56 Calculating K from DG o Try It!!! Calculate K p at 25 o C: 2H 2 O(l) 2H 2 (g) + O 2 (g) G fo kj Does your answer make sense? 56

57 Calculating DG o from K Calculate DG o for the aqueous dissolution of AgCl at 25 o C. K sp = 1.6 x AgCl(s) Ag + (aq) + Cl - (aq) Does your answer make sense? 57

58 One More!!! N 2 (g) + 3H 2 (g) 2NH 3 (g) P o (atm) What is DG at these initial pressures? Predict reaction direction. (K p = 6.59 x 10 5 at 25 o C) 58

59 The End 59

60 Warm-up What is the ph when 25 ml of 0.12 M HI is mixed with 32 ml of M Ba(OH) 2? 60

61 Warm-up Salt water is evaporated to dryness. What happens to the entropy of: Na + ions and Cl - ions? water? 61

62 Warm-up N 2 (g) + 3H 2 (g) 2NH 3 (g) S o (J/K) H o (kj) Under standard conditions, what temperature is this a spontaneous reaction? 62

63 Warm-up What is the sign of DG when the equilibrium constant is very small? Is the reaction spontaneous (Explain using the equation relating these two variables.) 63

64 Warm-up What is the difference between DG & DG o? For a given reaction, when does DG rxn = DG o? What is DG o and DS o when water condenses at 100 o C? (DH vap = 40.6 kj/mol) 64