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Outline: 2/2/07

Outline: 2/2/07. Chem Dept Seminar - today @ 4pm Exam 1 – two weeks from today…. Outline Quiz #3 Chapter 15 - Kinetics (cont’d): - determining reaction order - integrated rate law calcs. Quiz #3. Please put away all books and papers.

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Outline: 2/2/07

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  1. Outline: 2/2/07 • Chem Dept Seminar - today @ 4pm • Exam 1 – two weeks from today… • Outline • Quiz #3 • Chapter 15 - Kinetics (cont’d): • - determining reaction order • - integrated rate law calcs

  2. Quiz #3 • Please put away all books and papers. • If you don’t have a calculator, just set up the equations correctly…

  3. Quiz #3 • Please turn your quiz over and pass it to your right.

  4. Quiz #3 • Reaction was endothermic… = DH products - DHreactants Reaction increased entropy… = DS products - DSreactants • DG = DH - TDS • spontaneous when DG = 0 0 = DG + RT ln Q

  5. Quiz #3 • Reaction was endothermic… = DH products - DHreactants Reaction increased entropy… = DS products - DSreactants • DG = DH - TDS • spontaneous when DG = 0 0 = DG + RT ln Q

  6. R = 0.008315 kJ/mol K T = 298 K When do you use DG = DG + RT ln Q ? DGrxn = DGorxn+ RT ln Q Calculate DGreaction for the reaction @ 25°C: C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) if p=0.25 bar for each gaseous substance. DGorxn= -1325 kJ/mol (from appendix D)

  7. Q = Q = (0.25)2/(0.25)3 Q = 4 Problem 14.40 (c): What isQ? Calculate DGreaction for the reaction @ 25°C: C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) if p=0.25 bar for each gaseous substance. DGrxn = -1325 + 2.48 ln 4 = -1322

  8. How many problems of this type have you already worked out on your own.... • I have done dozens or more… • I have done many, but I still miss some… • I have done a couple from the book & CAPA • I have done the CAPA ones… • Never seen this type of problem before..

  9. Practice Problems: Chapter 14 • 14.11, 14.15, 14.17, 14.19, 14.23, 14.25, 14.27, 14.31, 14.35, 14.37, 14.38, 14.41, 14.43, 14.49, 14.51, 14.53, 14.55, 14.57, 14.61, 14.65, 14.67, 14.71, 14.75, 14.77, 14.79, 14.81, 14.91, 14.101, 14.103

  10. Back to Chapter 15…

  11. There are two forms to know: First order: ln[A] = ln[A]o-kt Second order: 1/[A] = 1/[A]o+kt 15-3 15-5 The rate equation cannot be predicted, it can only be measured empirically. • Calculate k from initial rates • Use the integrated form of the rate eqn. to solve for concentration (Section 15.4)

  12. Can use data to findkand reaction order How do you find the reaction order?

  13. Plot data

  14. ln[A] = ln[A]o-kt Plot both…. slope 1/[A] = 1/[A]o+kt Only one will be truly linear…. Rate = k [NO2]2

  15. Example: 1st or 2nd order decay?

  16. Try both plots…. Rate = k [cis-AB]

  17. Worksheet #4 Rate = k pNO2 pCl2 Know the “order” from the isolation experiments: e.g. double pNO and see what happens to rate… 0.000040 atm/s = k (0.1 atm)2 (0.1 atm) k = 0.040 1/atm2 s Don’t forget to practice lots of problems from the back of the chapter in the textbook

  18. There are two forms to know: First order: ln[A] = ln[A]o-kt Second order: 1/[A] = 1/[A]o+kt 15-3 15-5 Summary: (initial rates) • Can build a rate law from observed data: Rate = k [A]m[B]n • m, n depend only on the chemical reaction under consideration…. • Can use integrated rate laws to predict rates, concentrations at various times, etc.

  19. The integrated forms of the rate laws are important: Can predict [concentration] as a function of time! • Example: • The decomposition reaction of NO2 is second- order in [NO2], with a rate constant of 1.51 M-1s-1. • If the initial concentration is 0.041M, when will the concentration = 0.010 M?

  20. “Rate Law” of a reaction: • Rate =k[NO2]2 • The integrated solution: • 1/[NO2] - 1/[NO2]o = kt • 1/0.010 - 1/0.041 = 1.51M-1s-1t • t = ? • 50 sec

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