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# Ch16 z5e free energy

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### Ch16 z5e free energy

1. 1. Chapter 16 pp Spontaneity, entropy and free energy Note: For online HW, you might need to use the thermodynamic data sheet posted on my downloads website in the “helps” section if the values are not already in the problem. You need to use ∆G values that have at least one decimal place and your textbook only goes to the “one’s” place.
2. 2. Z5e 791 16.1 Methane and Oxygen React The products have lower potential energy than the reactants, resulting in energy flow (heat) to the surroundings.
3. 3. 16.1 Spontaneous A reaction that will occur without outside intervention. We can’t determine how fast. We need both thermodynamics and kinetics to describe a reaction completely. Thermodynamics compares initial &amp; final states ( i.e., potential energy). Kinetics describes pathway between (i.e., the reaction rate) .
4. 4. Z5e 792 Figure 16.2 Rate of Reaction Rate is a function of the pathway (kinetics) Spontaneity is a function of the potential energies (thermodynamics)
5. 5. Thermodynamics 1st Law - the energy of the universe is constant . Keeps track of thermodynamics, but doesn’t correctly predict spontaneity. Entropy (S) is disorder or randomness. 2nd Law - the entropy of the universe increases .
6. 6. Entropy Defined in terms of probability . Substances take the arrangement that is most likely. The most likely is the most random. Calculate the number of arrangements for a system to determine entropy.
7. 7. One particle: 2 possible arrangements (microstates) 50 % chance of finding the left empty
8. 8. Two particles: 4 possible microstates 25% chance of finding the left empty 50% chance of them being evenly dispersed
9. 9. Figure 16.4 Three Possible Arrangements (states) of Four Molecules in a Two-Bulbed Flask (Each arrangement has several microstates as seen in next slide)
10. 10. 3 possible arrangements (11 microstates - top one is duplicative) 8% chance of finding the left empty 50 % chance of them being evenly dispersed
11. 11. Gases Gases completely fill their chamber because there are many more ways to do that than to leave half empty. S solid &lt; S liquid &lt;&lt; S gas There are many more ways for the molecules to be arranged as a liquid than a solid ( positional entropy). Gases have a huge number of positions possible (more Entropy S).
12. 12. Positional Entropy For each pair below, choose the one with the higher positional entropy per mole ( i.e. , S is (+)) at a given temperature: Solid CO 2 and gaseous CO 2 . . . CO 2( g ) - more volume so more positions. N 2 gas at 1 atm vs. at 1.0 x 10 -2 atm (hint: Boyle’s Law) 1.0 x 10 -2 atm since Boyle’s law says lower P = higher V (more positions).
13. 13. 16.2 Entropy &amp; 2nd Law of Thermodynamics Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate from the solvent. 2nd Law of thermodynamics - the entropy of the universe is increasing  S univ =  S sys +  S surr If  S univ is (+) the process is spontaneous. If  S univ is (-) the process is spontaneous in the opposite direction. If  S univ is 0, then at equilibrium
14. 14. Cell metabolism In a living cell, complex molecules are assembled from simple ones. Is this consistent with the 2nd law? Yes. Only  S univ must be (+). We can have (-)  S sys as long as  S surr is both larger and positive
15. 15. For exo thermic processes  S surr is positive . For endothermic processes  S surr is negative . Consider this process H 2 O(l)  H 2 O(g)  S sys is (+); system gains entropy)  S surr is (-); surroundings lose entropy (molecular motion decreases)  S univ depends on temperature. pp
16. 16. 16.3 Temperature and Spontaneity Entropy changes in the surroundings are determined by the heat flow . An exo thermic process is favored because by giving up heat the entropy of the surroundings increases. The size of  S surr depends on temperature.  S surr = -  H/T (T in Kelvin)
17. 17.  S sys  S surr  S univ Spontaneous? - - - + + + + - ? Yes No, Reverse At low temp (  s surr magnitude &gt;  S sys At high temp (  s sys magnitude &gt;  S surr Z5e 803 Table 16.3 pp + - ?
18. 18. 16.4 Gibb&apos;s Free Energy G = H - TS (no subscript  “system”) Never used this way.  G =  H - T  S at constant temperature Divide both sides by -T -  G/T = -  H/T +  S -  G/T =  S surr +  S -  G/T =  S univ If  G is negative at constant T and P, the Process is spontaneous .
19. 19. Let’s Check For the reaction H 2 O(s)  H 2 O(l)  Sº = 22.1 J /K mol &amp;  Hº = 6.030 k J /mol Calculate  G &amp;  S univ at -10ºC, 0ºC &amp; 10ºC We use  G o =  H o - T  S o ( see next slide &amp; Table 16.4 p. 760 for results). We see that  G o = (+) at -10ºC (spontaneous in the opposite direction), zero at 0ºC (at =m) &amp; (-) at 10ºC (spontaneous as written).
20. 20. Z5e 804 Table 16.4 We see that  G o = (+) at -10ºC (spontaneous in the opposite direction), zero at 0ºC (at =m) &amp; (-) at 10ºC (spontaneous as written). So  G o = 0 at fp and mp (since at =m; use for preptest)
21. 21. Predicting Spontaneity Spontaneity can be predicted from the signs of  H and  S. See Table 16.5 p 761 (and next slide). You will have a test question on this table! Other hints especially for preptest and test: You can’t add kJ to J - convert On MC questions, watch what temperature units the answer is in. May need to convert K to Cº
22. 22.  G =  H - T  S pp + - At all Temp &amp; exotherm + + At high temp &amp; endoth., “ entropy driven” - - At low temperatures, “ enthalpy driven” + - Not at any temperature, Reverse is spontaneous  H  S Spontaneous?
23. 23. 16.5 Entropy Changes in Chem. Rxns pp Predict the Sign of  Sº for the following: Thermal decomposition of solid calcium carbonate. . . Hint: write reaction and examine positional entropy. CaCO 3(s) CaO (s) + CO 2(g) Positional entropy increases since solid to gas so S = (+) Oxidation of SO 2 in air. . . 2SO 2(g) + O 2(g) 2SO 3(g) 3 molecules of gas become 2 so S is (-).
24. 24. Third Law of Thermo The entropy of a pure crystal at 0 K is 0. This gives us a starting point. All other substances must have S &gt; 0. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. ∑ Products - ∑reactants to find  Sº (a state function) More complex molecules - higher Sº.
25. 25. Z5e 810 Figure 16.6 H 2 O Molecule H 2 O molecule can vibrate and rotate in several ways as shown here. This freedom of motion leads to higher entropy for water than for a substance like hydrogen (a simple diatomic molecule with fewer possible motions).
26. 26. 16.6 Free Energy in Reactions  Gº = standard free energy change. This is the free energy change that will occur if reactants in their standard state turn to products in their standard state. Can’t be measured directly, can be calculated from other measurements.  Gº =  Hº - T  Sº Watch units of H (in k J) and S (in J). Change temp. to Kelvin. Use Hess’ Law with known reactions.
27. 27. Calculations Involving Energy If 2 mol reactants  4 mol product + 500 kJ, is the rxn likely spontaneous? Yes What is the sign of ∆S rxn when molten wax hardens? (-) How would your calculations change if the coefficients were not 1? ∆ H f and S values in  Gº =  Hº - T  Sº multiplied by the coefficients.
28. 28. Free Energy in Reactions pp There are tables of  Gº f (pp. A21 ff )  Gº rxn = ∑products - ∑reactants because it is a state function. Std. free energy of formation (  Gº f ) for any element in standard state is 0 (don’t forget!). Remember: Spontaneity tells us nothing about rate. Watch for trick questions that give you values of combustion (may need to change the sign - write the reaction to find out)
29. 29. Free Energy in Reactions pp Three methods to determine G o #1  Gº =  Hº - T  Sº Use  Hº = ∑n p  Hº f(products) - ∑n r  Hº f(reactants) Use  Sº = ∑n p  Sº products - ∑n r  Sº reactants find standard values in pp. A21 ff #2 Hess’ Law (free energy is state function) #3  Gº = ∑n p  Gº f(products) - ∑n r  Gº f(reactants) find standard values in pp. A21 ff remember:  Gº of element = 0
30. 30. Three methods to determine G o pp #1  Gº =  Hº - T  Sº @25ºC &amp; 1 atm 2SO 2(g) + O 2(g) 2SO 3(g) ∆Hº f (kJ/mol) -297 0 -396 Sº (J/K•mol) 248 205 257  Hº = ∑n p  Hº f(products) - ∑n r  Hº f(reactants)  Hº = [2(-396)] - [2(-297) + 0] = -198 kJ  Sº = ∑n p  Sº products - ∑n r  Sº reactants  Sº = [2(257)] - [2(248) + 205] = -187 J (expect (-) since 3 moles gas 2 moles)  Gº = -198 kJ -(298 K )(-187 J/K )(1 kJ /1000 J )  Gº = -142 kJ (spontaneous since (-))
31. 31. Three methods to determine G o pp #2 Hess’ Law (free energy is state function) C diamond(s) + O 2(g) CO 2(g)  Gº = -397 kJ C graphite(s) + O 2(g) CO 2(g)  Gº = -394 kJ Calculate  Gº for C diamond(s) C graphite(s) Steps . . . Reverse the 2nd reaction (change the sign) and add.  Gº = -397 kJ + 394 kJ = -3 kJ, but slow. Diamond is kinetically stable, but thermodynamically unstable.
32. 32. Three methods to determine G o pp #3  Gº = ∑n p  Gº f(products) - ∑n r  Gº f(reactants) 2CH 3 OH (g) + 3O 2(g) 2CO 2(g) + 4H 2 O (g)  Gº f(kJ/mol) -163 0 -394 -229 ∆ Gº = [2(-394) + 4(-229)] - [2(-163)] ∆ Gº = -1378 kJ/mol Large magnitude &amp; (-) sign means this is very favorable thermodynamically.
33. 33. Z5e 16.7: So Far . . . pp Three methods to determine ∆G o #1  Gº =  Hº - T  Sº Use  Hº = ∑n p  Hº f(products) - ∑n r  Hº f(reactants) Use  Sº = ∑n p  Sº products - ∑n r  Sº reactants find standard values in pp. A21 ff #2 Hess’ Law (free energy is state function) #3  Gº = ∑n p  Gº f(products) - ∑n r  Gº f(reactants) find standard values in pp. A21 ff remember:  Gº of element = 0 Now, let’s look at free energy &amp; pressure, then free energy &amp; equilibrium ( 2 more methods ).
34. 34. 16.7 Free energy and Pressure pp  G =  Gº + RT ln(Q) where Q is the reaction quotients (P products ÷ P reactants ) and follows the law of mass action. R = 8.314 J /Kmol (but ∆Gº is in k J!)
35. 35. 16.7 Free energy and Pressure pp  G =  Gº + RT ln(Q) Given: CO(g) + 2H 2 (g)  CH 3 OH(l) Would the reaction be spontaneous at 25ºC with P H 2 = 3.0 atm &amp; P CO = 5.0 atm?   Gº f CH 3 OH(l) = -166 kJ  Gº f CO(g) = -137 kJ  Gº f H 2 (g) = 0 kJ Which method (we now know 4) should we use? . . . Anytime you have a problem involving a gas and pressure you need an equation with “ R ”. So use  G =  Gº + RT ln(Q)
36. 36. Free energy and Pressure pp CO(g) + 2H 2 (g)  CH 3 OH(l)  Gº f CH 3 OH(l) = -166 kJ  Gº f CO(g) = -137 kJ  Gº f H 2 (g) = 0 kJ  G =  Gº + RT ln(Q) 1st, calc.  G º from standard free energies of formation (see pp. A21 ff or above) . . .  G º = -29 kJ [-166 k J] - [-137 k J + 2•0 k J] Convert kJ to J ( R is in joules ) = -2.9 x 10 4 J
37. 37. Free energy and Pressure pp CO(g) + 2H 2 (g)  CH 3 OH(l)  Gº = -2.9 x 10 4 J, P H2 = 3.0 atm &amp; P CO = 5.0 atm  Gº = -2.9 x 10 4 J/mol•rxn (from last slide) 2nd, calc.  G using  G =  Gº + RT ln(Q), but first have to calculate Q . . . Note: methanol not used in calculating Q. Why? . . . Pure liquid, so assign value = 1 , Q = . . .? Q = 2.2 x 10 -2 1 /[(5.0 atm)(3.0 atm) 2 ]
38. 38. Free energy and Pressure pp CO(g) + 2H 2 (g)  CH 3 OH(l)   Gº = -2.9 x 10 4 J, P H2 = 3.0 atm &amp; P CO = 5.0 atm @25º C  G º = -2.9 x 10 4 J/mol•rxn Q = 2.2 x 10 -2 1/[(5.0 atm)(3.0 atm) 2 ] So,  G =  Gº + RT ln(Q) = ? . . .  G = -38 kJ/mol Calculation is . . . -2.9 x 10 4 J/mol•rxn + (8.3145 J/mol)(25 + 273 K)•ln(2.2 x 10 -2 ) Is it spontaneous? . . . Yes.  G is negative .
39. 39. 16.8 Free Energy &amp; Equilibrium pp  G =  Gº + RT ln(Q)  G tells us spontaneity at current conditions. But, when will it stop? It will go to the lowest possible free energy, which may be an equilibrium. At equilibrium  G = 0, Q = K, so . . . 0 = ∆Gº + RT ln K So,  Gº = - RT lnK We now have 5 equations for ∆G ! You know to use this equation when you have a problem that has G &amp; K
40. 40. 5 Methods to Determine Gº pp #1  Gº =  Hº - T  Sº Use  Hº = ∑n p  Hº f(products) - ∑n r  Hº f(reactants) Use  Sº = ∑n p  Sº products - ∑n r  Sº reactants find standard values in pp. A21 ff #2 Hess’ Law (free energy is state function) #3  Gº = ∑n p  Gº f(products) - ∑n r  Gº f(reactants) find standard values in pp. A21 ff  Gº element = 0 #4 ∆G = ∆Gº + RT ln(Q) #5 ∆Gº = - RT ln(K)
41. 41. Figure 16.7 Balls Rolling Down into Two Types of Hills Goes to pure products since B is lowest free energy ( no equilibrium ) Doesn’t go to pure products ( i.e., there is an equilibrium ) since intermediate C has lowest G.
42. 42. Z5e 820 16.8 The Dependence of Free Energy on Partial Pressure A (g) B (g) Initial free energies as A goes to B Free energy of A lessens &amp; free energy of B increases Finally, partial pressures are equal and equilibrium is reached
43. 43. Figure 16.9 Free Energy and Equilibrium a.  Gº to reach =m, beginning with 1.0 mol A (g ) &amp; P A = 2.0 atm b.  Gº to reach =m, beginning with 1.0 mol B (g) &amp; P B = 2.0 atm c. G profile for A (g) B (g) with 1.0 mole of each at P total = 2.0 atm. Each point on the curve corresponds to the total free energy of the system for a given combination of A and B.
44. 44.   Gº K ∆Gº = - RT ln(K) = 0 &lt; 0 &gt; 0 = 1 &gt; 1 &lt; 1 pp
45. 45. Free energy and equilibrium pp N 2(g) + 3H 2(g) 2NH 3(g) P a 1.47 atm .0100 atm 1.00 atm P b 1.00 atm 1.00 atm 1.00 atm ∆Gº = -33.3 kJ/mol of N 2 consumed @ 25º C. Predict shift in =m for the above 2 cases. What to use? . . . Look for an equation with ∆G &amp; Q (since want initial conditions), calc. ∆G, if (-) then shifts to right (spontaneous). . . ∆ G = ∆Gº + RT ln(Q)
46. 46. Free energy and equilibrium pp N 2(g) + 3H 2(g) 2NH 3(g) P a 1.47 atm .0100 atm 1.00 atm P b 1.00 atm 1.00 atm 1.00 atm ∆Gº = -33.3 kJ/mol of N 2 consumed @ 25º C. ∆ G = ∆Gº + RT ln(Q). For P a : ∆G = . . .? ∆ G = 0 (did you convert J &amp; kJ?) (-3.33 x 10 4 J /mol) + (8.3145 J/K•mol)(298 K)ln(6.8 x 10 5 ) Be sure to use extra ( ) for ln ( 6.8 x 10 5 ) in your calculator! Do sig figs before adding. So, P a is at =m &amp; no shift occurs.
47. 47. Free energy and equilibrium pp N 2(g) + 3H 2(g) 2NH 3(g) P a 1.47 atm .0100 atm 1.00 atm P b 1.00 atm 1.00 atm 1.00 atm ∆Gº = -33.3 kJ/mol of N 2 consumed . ∆ G = ∆Gº + RT ln(Q) For P b : everything is at standard state . ∆ G = ∆Gº + RT ln(1) = ∆Gº = -33.3kJ/mol - ∆Gº means spontaneous, so products have lower free energy than reactants. So, shifts to right and K &gt; 1.
48. 48. Free energy and equilibrium pp 4Fe (s) + 3O 2(g) 2Fe 2 O ∆Hº f (kJ/mol) 0 0 -826 Sº f (J/K•mol) 27 205 90 Calc K @ 25ºC &amp; 1 atm. Which equation? Use ∆Gº = - RT ln(K) (std state) find K. First calculate ∆Gº from what??? . . . ∆ Gº = ∆Hº - T∆Sº Calc ∆Hº rxn &amp; Sº rxn Your answer for ∆Gº is . . . ∆ Gº = -1.490 x 10 6 J (did you convert kJ?) Calculations on next slide.
49. 49. Free energy and equilibrium pp 4Fe (s) + 3O 2(g) 2Fe 2 O ∆Hº f (kJ/mol) 0 0 -826 Sº (J/K•mol) 27 205 90 ∆ Gº = ∆Hº - T∆Sº Calc ∆Hº rxn &amp; Sº rxn ∆ Hº rxn = {[2(-826 kJ)] - [0]} x 1000 J/kJ = -1.652 x 10 6 J Sº rxn = [(2)(90)] - [(4)(27) + (3)(205)] = -545 J/K ∆ Gº = -1.652 x 10 6 J - ( 25 + 273 K)( -545 J/K ) = . . . As we saw, ∆Gº = -1.490 x 10 6 J (Remember to convert kJ to J and ºC to K) Now, find K. . . (next slide)
50. 50. Free energy and equilibrium pp 4Fe (s) + 3O 2(g) 2Fe 2 O ∆Hº f (kJ/mol) 0 0 -826 Sº (J/K•mol) 27 205 90 (What is K at 25ºC &amp; 1 atm?) ∆Gº = -1.490 x 10 6 J Plug into ∆Gº = - RT ln(K) &amp; find K . . .? ln K = 601, so K = e 601 = 10 261 Even though a humongous K = 10 261 (too large for my calculator) this means iron rusting is thermodynamically favored. This does not tell us the rate of iron rusting.
51. 51. Temperature dependence of K pp  Gº= - RT lnK =  Hº - T  Sº ln(K) = -  Hº/R(1/T)+  Sº/R i.e., y = mx + b Get a straight line of lnK vs. 1/T with slope of  Hº/R So, if we experimentally determine K at different temperatures, can graphically get  Hº and  Sº
52. 52. 16.9 Free energy And Work Free energy is the energy available to do work . It represents the maximum amount of work possible at a given temperature and pressure. Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.
53. 53. Figure 16.10 A Battery A battery can do work by sending current to a starter motor. It can be recharged by forcing current through it in the opposite direction.
54. 54. Figure 16.10 A Battery If the current flow in both processes is infinitesimally small, w1 = w2 and the process is reversible
55. 55. Figure 16.10 A Battery But, in real world, current flow is finite , so w2 &gt; w1 and process is irreversible ; that is, the universe is different after the process occurs.
56. 56. Figure 16.10 A Battery All real processes are irreversible.
57. 57. Reversible v. Irreversible Processes pp Reversible : The universe is exactly the same as it was before the cyclic process. Irreversible : The universe is different after the cyclic process. All real processes are irreversible -- (some work is changed to heat).