1. Ppt04 (PS8), Thermodynamics energy changes (heat flows) Very global (“the universe”) Applies to all fields of science Led to creation of heat engines/refrigerators/air conditioners In chemistry, focus is on physical and chemical changes How can we use “spontaneous processes” to “do work”? We can use tabulated data to make predictions (and even calculate K’s for possible processes!) 1Ppt08

2. “Spontaneous”  “fast” !! Spontaneity refers to “directionality” –Does it occur (on its own) in the forward [or stated] direction, or does it not? Thermodynamics will answer / address this question Does not depend on how the process is carried out. Fast refers to rate at which it occurs (in the spontaneous direction) –How fast will it occur? Kinetics will answer / address this question Does depend on how the process occurs (“pathway”; transition state, activation energy, etc.) Recall that we should be spontaneously combusting right now!! 2Ppt08

3. Figure 17.4 Thermodynamics domain vs. Kinetics domain “Pathway dependent” THERMODYNAMICS Initial and final states, spontaneity “Path independent” 3Ppt08

4. Reminder: 1 st Law of Thermodynamics The total amount of energy in the universe never changes (it’s constant)  E univ = 0   E sys = -  E surr “If energy leaves the system, it must go to the surroundings, and vice versa” Has nothing to do with “spontaneity” –1 st Law is consistent with a casserole dish coming out of the oven colder than when it was put in, as long as the oven would get hotter Ppt084

5. Spontaneity and the 2 nd Law A spontaneous process is one that occurs (at a specified set of conditions) “without external intervention”. –Gases expand into the volume of their containers –Ice will melt at 10°C –Water will freeze at -10°C –A chemical reaction will occur in the forward direction if Q < K Ppt085

6. Spontaneity and the 2 nd Law (continued) At the same conditions, the reverse process of a spontaneous one is not spontaneous –Gases won’t spontaneously “contract” into a smaller volume (perfume molecules going back into the bottle after opened?) –Ice will not freeze at 10°C –Water will not melt at -10°C –A chemical reaction will not occur in the reverse direction if Q < K Ppt086

7. Spontaneity and the 2 nd Law (continued) What determines whether a given process is spontaneous or not? –This question is not addressed by the 1 st Law; it is addressed by the 2 nd Law. 2 nd Law: For a spontaneous process to occur, the entropy (S) of the universe must increase.  S univ > 0 for any spontaneous process Ppt087

8. **  S univ   S sys !** Be careful! The universe is made up of two “parts”— system and surroundings: ΔS univ = ΔS sys + ΔS surr It doesn’t ultimately matter if ΔS sys is positive, or if ΔS surr is positive. The key is whether ΔS univ is positive! (next slide). Ppt088

9. Interplay of ΔS sys and ΔS surr in Determining the Sign of ΔS univ (Table 16.3, Zumdahl) This turns out to be very temperature dependent (See Section 17.4 in Tro). We will come back to this issue later. See next slide plus equals ΔS sys + ΔS surr = ΔS univ 9Ppt08

10. Bar plots to visualize idea on prior slide (system and surroundings both contribute to  S univ !) NOTE: These two examples both have  S sys 0, but all possible variations are possible! ΔS sys + ΔS surr = ΔS univ |ΔS surr | > |ΔS sys ||ΔS sys | > |ΔS surr | 10Ppt08

11. What is entropy? Hard thing to define conceptually! –Something like “energy dispersal”--not necessarily “spatial” dispersal, but energy dispersed over the various motions of atoms & molecules. Reflects quality or usefulness of energy. Let’s start with individual substances –Look at what affects the amount of entropy in a sample of a single substance. Then we’ll look at mixtures of substances (in a reactive system, e.g.) –When a change occurs in the system, S sys changes because there are new substances present (or new physical states) We’ll consider the surroundings last –Just a “bunch of substances not doing anything” (Prof. Mines’ view) 11Ppt08

12. Patterns for Entropies of Substances (see handout/outline for details; Tro does this later; Section 17.6) Entropy of a sample of a substance increases with: Recall that we’ll typically only specify substances that undergo some change to be part of the system. –Energy added (T increase or phase change) Solid < Liquid << Gases –Volume (gas or solution species) –Complexity of basic unit of substance –Number of moles of the substance Entropy, like energy, is a “per mole” kind of quantity 12Ppt08

13. Differences in entropy of physical states (assume a given T) 13Ppt08

14. Fig. 17.5 Entropy of a substance increases with T, and depends (significantly) on state (s, l, g) 14Ppt08

15. One “way” to arrange the energy (one [micro]state) Two “ways” to arrange the energy (two [micro]states) E = 5 J Both systems have 4 J of energy, but the entropy of System B is greater because it has two ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see Slide 17)! Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E System A (4 J)System B (4 J) 15Ppt08 What if an energy state at 5 J was present? Would the # of microstates change in either system?  No, because there isn’t enough energy (in either system) to access that 5 J state. But…

16. What if 2-3 more units of energy were added (to make a total of 6-7 J)? Two “ways” to arrange the energy (vs. one before); two[micro]states Three “ways” to arrange the energy (vs two before); three [micro]states E = 5 J In both cases, adding energy (raising T) leads to increased S Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E System A (6-7 J)System B (6-7 J) Both systems have 6-7 J of energy, but the entropy of System B is (still) greater because it has three ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see next slide)! 16Ppt08 E E = 5 J

17. More energy levels in gas! 17Ppt08

18. Patterns for Entropies of Substances (see handout/outline for details; Tro does this later; Section 17.6) Entropy of a sample of a substance increases with: –Energy added (T increase or phase change) Solid < Liquid << Gases –Volume (gas or solution species) –Complexity of basic unit of substance –Number of moles of the substance Entropy, like energy, is a “per mole” kind of quantity 18Ppt08 Adding units of energy Adding more accessible energy states

19. (NOTE: Appendix has lots more data. Use appendix for PS8!) 19Ppt08

20. (From another text [McMurry]) 20Ppt08

21. Predicting whether processes are “entropy increasing” or “decreasing” (in the system) (the most dominant effect in most cases is the number of moles of gases made and lost [  n gas ], not the complexity of the substance[s]): C(s) + 2 H 2 (g)  CH 4 (g) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) [see next slide] 2 CO(g) + O 2 (g)  2 CO 2 (g) CaCO 3 (s)  CaO(s) + CO 2 (g) 2 NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 21Ppt08

22. Can quantify the change in S (at a given T) using standard entropies of substances C(s) + 2 H 2 (g)  CH 4 (g)  S rxn ° = 186.2 – (2.4 + 2 x 130.6) 2.4 2 x 130.6 186.2 = -77.4 J/K (per mol of C reacted) (consistent with prediction, S decreases) –Although CH4 is more complex than H 2, it doesn’t have twice the entropy (per mole) as H 2, so the dominant effect here (as usual) is the change in the number of moles of gases on reaction,  n gas ) N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  S rxn ° = 2 x 192.3 – (191.5 + 3 x 130.6) 191.5 3 x 130.6 2 x192.3 = -198.7 J/K (per mol of N 2 ) –Again, consistent with prediction (S decreases). Though NH3 more complex, only 2 moles of it form compared to 4 moles of gas “lost”). 22Ppt08

23. The entropy of a perfect crystal at 0 K is zero (3 rd Law). Tro, Fig. 17.8 Zumdahl, Fig 16.5 23Ppt08 W = “the number of microstates” (ways to arrange energy)

24. Entropy increases for a substance as energy is added to it (either T increases, or a phase change occurs) Curve differs for different substances. This is how standard entropies of substances are determined (at, say, 298 K) 24Ppt08

25. Relating Entropy of the Surroundings to a Property of the System--What is “free energy” (G)? See Section IV of Handout Outline 25Ppt08

26. © 2011 Pearson Education, Inc. EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from  H and  S SOLUTION Consider the reaction for the decomposition of carbon tetrachloride gas: (a) Calculate  G at 25  C and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25  C, determine at what temperature (if any) the reaction becomes spontaneous. The reaction is not spontaneous. At the specific conditions / concentrations of the reaction system! Consider only part (a) for now. 26Ppt08 (a) Use Equation 17.9 to calculate  G from the given values of  H and  S. The temperature must be in kelvins. Be sure to express both  H and  S in the same units (usually joules).

27. RECALL: Interplay of ΔS sys and ΔS surr in Determining the Sign of ΔS univ (Table 16.3, Zumdahl) This turns out to be very temperature dependent (See Section 17.4 in Tro). We will come back to this issue later. See next slide plus equals ΔS sys + ΔS surr = ΔS univ 27Ppt08

28. 28

29. Revisiting Earlier Example If both of these plots represent the same exact process under the same exact conditions EXCEPT THAT ONE IS AT A HIGHER TEMPERATURE, then which plot represents the higher T and which the lower T? How do you know? What would the bar plot look like if the T went to infinity?To zero? 29Ppt08 (see next slides →)

30. As T → , |  S surr | → 0! (so  S univ →  S sys ) T increasing T~  30Ppt08

31. As T → 0, |  S surr | →  ! (so  S univ →  S surr ) T decreasing T very small ( Y-axis scale much bigger (zoomed out);  S sys same as on right.) 31Ppt08

32. Summary, T-dependence of  S univ At very high T,  S univ approaches  S sys –Because  S surr becomes tiny (and  S sys remains essentially unchanged) At very low T,  S univ approaches  S surr –Because  S surr becomes huge, while  S sys remains essentially unchanged 32 NOTE: Many people end up taking a different approach to this “T-dependence” issue: They look at  G sys rather than  S univ. We’ll explore this next! **A more detailed summary table of this will be discussed a bit later. Ppt08

33. Thus: As T goes to… 33 …infinity,  G →  G =  H –T  S …zero,  G →  H (  S surr dominates) -T  S (  S sys dominates) Ppt08

34. Tro’s “version” of this 34 Thus: As T goes to… …infinity,  G →  G =  H –T  S …zero,  G → HH -T  S HH  G sys “view” NOTE: You could also look at each of these from a  S univ “view”! Ppt08

35. “Global” (entropy) view (Review of earlier slide) 35 Thus: As T goes to… …infinity,  S univ →  S univ =  S sys +  S surr ;  S surr = …zero,  S univ →  S surr  S sys  S surr  S univ “view”  S surr  S sys        S surr > 0)  S surr < 0)  S sys < 0)  S sys > 0)  Ppt08

36. 36 Table 16.4 (Zumdahl) Results of the Calculation of ΔS univ and ΔG° for the Process H 2 O (s)  H 2 O (l) at -10°C, 0°C, and 10°C T-Dependence Explored Further Ppt08

37. © 2011 Pearson Education, Inc. FOR PRACTICE 17.3 Consider the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)∆H = –137.5 kJ; ∆S = –120.5 J/K Calculate  G at 25  C and determine whether the reaction is spontaneous. Does  G become more negative or more positive as the temperature increases? EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from  H and  S (b) Since  S is positive,  G will become more negative with increasing temperature. To determine the temperature at which the reaction becomes spontaneous, use Equation 17.9 to find the temperature at which  G changes from positive to negative (set  G = 0 and solve for T). The reaction is spontaneous above this temperature. (a) Use Equation 17.9 to calculate  G from the given values of  H and  S. The temperature must be in kelvins. Be sure to express both  H and  S in the same units (usually joules). SOLUTION Consider the reaction for the decomposition of carbon tetrachloride gas: (a) Calculate  G at 25  C and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25  C, determine at what temperature (if any) the reaction becomes spontaneous. The reaction is not spontaneous. At the specific conditions / concentrations of the reaction system! Now consider part (b) 37Ppt08

38. Relationship of  G to Q (and  G  ) See Outline (and board) ΔG = ΔG° + RT ln Q 38Ppt08

39. Relationship of  G  to K See Outline—Derive from prior slide’s relationship! 39Ppt08

40. Table 16.6 (Zumdahl) Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction 40  G  = -RT lnK Ppt08

41. 41Ppt08

42. From another text These data provide a more direct way to calculate the  G  for a chemical reaction equation (i.e., you don’t need to calculate  H  and  S  if you know  G  f values! 42Ppt08

43. Fig. 17.10 (Tro). Why “free” energy? 43  H  = -74.6 kJ  S  = -80.8 J/K  G  = -50.5 kJ Ppt08

44. Table 16.1 The Microstates That Give a Particular Arrangement (State) Ppt0844

45. Table 16.2 Probability of Finding All the Molecules in the Left Bulb as a Function of the Total Number of Molecules Ppt0845