1. 6-1 Unlike  H,  E, or  S…  G depends on amounts of reactants and products present………This is a parameter characterized by the letter Q (reaction quotient) for solutions: quantities involve concentrations for gasses, quantities involve pressures pure solids, and liquids always contribute a factor of 1 to Q for solutions:for gasses: if, say A is a pure solid, C is a pure liquid, and B and D are gasses So what???

2. 6-2  G =  G o + RT ln Q Under standard conditions all liquids are pure liquids, all solids are pure solids, all solutions are solutions have 1M concentrations all gasses have 1atm pressures- therefore Q = 1 and lnQ = 0 so  G  G o Relationship of  G and Reaction Quotient Two interpretations of  G o 1. free energy change per mol of reaction when reactants in their standard states convert to products in their standard states 2. free energy available to a system under standard conditions i.e. the maximum amount of work a system can do on the surroundings under standard of conditions Recall that  G is the maximum amount of work a system can do under a given set of conditions. Also, a system at equillibrium cannot do work

3. 6-3  G =  G o + RT ln Q Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0 so  G  G o At equilibrium,  G=0 so K is equilibrium constant, a particular Q that describes amounts of reactants and products present in a system at equilibrium

4. 6-4 Free Energy, Equilibrium and Reaction Direction If Q/K, then ln Q/K < 0; the reaction proceeds to the right (  G < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (  G > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (  G = 0)

5. 6-5 Figure 20.12 The relation between free energy and the extent of reaction  G 0 1  G 0 > 0 K <1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Q=1 also  G o G of remaining reactant +G of product at Q=1

6. 6-6 When  G o =0=  H o -T  S o at equillibrium when Q=1, standard cond.  H o /T=  S o Q=1=K G reactants G products

7. 6-7 FORWARD REACTION REVERSE REACTION Table 20.2 The Relationship Between  G 0 and K at 25 0 C  G 0 (kJ) KSignificance 200 100 50 10 1 0 -10 -50 -100 -200 9x10 -36 3x10 -18 2x10 -9 2x10 -2 7x10 -1 1 1.5 5x10 1 6x10 8 3x10 17 1x10 35 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction

8. 6-8 Sample Problem 20.7: PROBLEM: Calculating  G at Nonstandard Conditions The oxidation of SO 2, which we considered in Sample Problem 20.6 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298K and at 973K. (  G 0 298 = -141.6kJ/mol of reaction as written using  H 0 and  S 0 values at 973K.  G 0 973 = -12.12kJ/mol of reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO 2, 0.0100atm of O 2, and 0.100atm of SO 3 and kept at 25 0 C and at 700. 0 C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate  G for the system in part (b) at each temperature. PLAN:Use the equations and conditions found on slide.  G o =  H o -T  S o =[2(-396)-2(-296)-0]-T[2(.25666)-2(.24841)-(.205)] why not  G o from table?

9. 6-9 SOLUTION: Sample Problem 20.7: Calculating  G at Nonstandard Conditions continued (2 of 3) (a) Calculating K at the two temperatures:  G 0 = -RTlnK so At 298, the exponent is -  G 0 /RT = - (-141.6kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(298K) = 57.2 = e 57.2 = 7x10 24 At 973, the exponent is -  G 0 /RT (-12.12kJ/mol)(10 3 J/kJ) (8.314J/mol*K)(973K) = 1.50 = e 1.50 = 4.5

10. 6-10 Sample Problem 20.7: Calculating  G at Nonstandard Conditions continued (3 of 3) (b) The value of Q = pSO 3 2 (pSO 2 ) 2 (pO 2 ) = (0.100) 2 (0.500) 2 (0.0100) = 4.00 Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight. (c) The nonstandard  G is calculated using  G =  G 0 + RTlnQ  G 298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(298K)(ln4.00)  G 298 = -138.2kJ/mol  G 973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/10 3 J)(973K)(ln4.00)  G 298 = -0.9kJ/mol

11. 6-11 Okay, so at what temperature is this reaction at equillibrium under standard conditions? What would happen if  H and  S had different signs?

12. 6-12 A slightly more mundane example: normal boiling point of water 373.15 K (liquid and vapor in eq.) I.e. K=Q=P vap =1,  G=  G o =0 find equillibrium vapor pressure of water at 50 C what if Pvap is 0.05 atm? what if Pvap is 0.12 atm?

13. 6-13 okay then look familiar?

14. 6-14 Another example: CH 3 OH (g) ---> CO (g) + 2 H 2(g) What is K at 25 C?  G o =  H o -T  S o =[(-110.5)+2(0) - (-201.2)]-(298.15)[(.1975)+2(.1306)-(.238)] = 90.7 - (298.15)(.2207) = 24.9 kJ/mol another way: what if P H2 =1.5 atm, P CO =.21 atm, P CH3OH =0.0075 atm what about K at 250 C?

15. 6-15 Can’t use Gibbs free energies of formation, why?  G o (250)= 90.7 - (523.15)(.2207)= -24.8 KJ/mol what if P H2 =0.012 atm, P CO =.023 atm, P CH3OH =0.0055 atm Q=0.000601 < K

16. 6-16 Questions of whether a reaction is spontaneous or not are irrelevant to the question of how fast they will occur diamond will spontaneously convert to graphite under standard conditions C(diamond) ---> C(graphite)  G=  G o +RT lnQ =  G o +0=  H o =T  S o = -2.866 kJ/mol Yet, you still shell out big bucks for Diamonds

17. 6-17 Chapter 16 Kinetics: Rates and Mechanisms of Chemical Reactions

18. 6-18 Kinetics: Rates and Mechanisms of Chemical Reactions 16.1 Factors That Influence Reaction Rates 16.2 Expressing the Reaction Rate 16.3 The Rate Law and Its Components 16.4 Integrated Rate Laws: Concentration Changes over Time 16.5 Reaction Mechanisms: Steps in the Overall Reaction 16.6 Catalysis: Speeding Up a Chemical Reaction

19. 6-19 Figure 16.1 Reaction rate: the central focus of chemical kinetics Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

20. 6-20 Time (s) Concentration of O 3 (mol/L) 0.0 20.0 30.0 40.0 50.0 60.0 10.0 3.20x10 -5 2.42x10 -5 1.95x10 -5 1.63x10 -5 1.40x10 -5 1.23x10 -5 1.10x10 -5 Table 16.1 Concentration of O 3 at Various Time in its Reaction with C 2 H 4 at 303K Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

21. 6-21 Figure 16.5 The concentrations of O 3 vs. time during its reaction with C 2 H 4

22. 6-22 Figure 16.6 Plots of [C 2 H 4 ] and [O 3 ] vs. time Tools of the Laboratory

23. 6-23 Sample Problem 16.1 PLAN: SOLUTION: Expressing Rate in Terms of Changes in Concentration with Time PROBLEM:Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may be used by Earth-bound engines in the near future. 2H 2 ( g ) + O 2 ( g ) 2H 2 O( g ) (a) Express the rate in terms of changes in [H 2 ], [O 2 ], and [H 2 O] with time. (b) When [O 2 ] is decreasing at 0.23 mol/L*s, at what rate is [H 2 O] increasing? Choose [O 2 ] as a point of reference since its coefficient is 1. For every molecule of O 2 which disappears, 2 molecules of H 2 disappear and 2 molecules of H 2 O appear, so [O 2 ] is disappearing at half the rate of change of H 2 and H 2 O. - 1 2  [H 2 ] tt = -  [O 2 ] tt = +  [H 2 O] tt 1 2 0.23mol/L*s = +  [H 2 O] tt 1 2 ; = 0.46mol/L*s  [H 2 O] tt rate = (a)  [O 2 ] tt -= -(b)

24. 6-24 Sample Problem 16.2 SOLUTION: Determining Reaction Order from Rate Laws PROBLEM:For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. (a) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ); rate = k[NO] 2 [O 2 ] (b) CH 3 CHO( g ) CH 4 ( g ) + CO( g ); rate = k[CH 3 CHO] 3/2 (c) H 2 O 2 ( aq ) + 3 I - ( aq ) + 2H + ( aq ) I 3 - ( aq ) + 2H 2 O( l ); rate = k[H 2 O 2 ][ I - ] PLAN:Look at the rate law and not the coefficients of the chemical reaction. (a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall. (b) The reaction is 3/2 order in CH 3 CHO and 3/2 order overall. (c) The reaction is 1st order in H 2 O 2, 1st order in I - and zero order in H +, while being 2nd order overall.

25. 6-25 Figure 16.4 Collision energy and reaction rate Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.