Presentation Transcript

1. Chemistry 102(01) Spring 2013 Instructor: Dr. UpaliSiriwardane e-mail: upali@coes.latech.edu Office: CTH 311 Phone257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F9:30 - 11:30 am.  Test Dates: September 27, 2013 (Test 1): Chapter 12 & 13 April 24, 2013 (Test 2): Chapter 14 & 15 May13, 2013 (Test 3) Chapter 16 & 17 May 15, 2012 (Make-up test) comprehensive: Chapters 12-17 9:30-10:45:15 AM, CTH 328

2. Chapter 13. Chemical Kinetics

   13.1 Catching Lizards 563 13.2 The Rate of a Chemical Reaction 564 13.3 The Rate Law: The Effect of Concentration on Reaction Rate 569 13.4 The Integrated Rate Law: The Dependence of Concentration on Time 573 13.5 The Effect of Temperature on Reaction Rate 581 13.6 Reaction Mechanisms 588 13.7 Catalysis 593
3. Chemical Kinetics Definitions and Concepts a) rate of reations b) rate law b) rate constant c) order d) differential rate law c) integral rate law d) Half-life law
4. Every chemical reaction has a Rate Law The rate law is an expression that relates the rate of a chemical reaction to a constant (rate constant-k) and concentration of reactants raised to a power. The power of a concentration is called the order with respect to a particular reactant. Rate Law
5. Rate Law E.g. aA+ bB-----> cC rate a [A]l[B]m rate =-1/a d[A]/dt= k [A]l[B]m; k = rate constant [A] = concentration of A [B] = concentration of B l = order with respect to A m = order with respect to B l & m have nothing to do with stoichiometric coefficients
6. Differential Rate Law E.g. 2 N2O5(g) -----> 4 NO2 (g) + O2 (g) rate= - ½ d[N2O5]/dta[N2O5]1 rate = - ½ d[N2O5]/dt= k [N2O5]1 k = rate constant [N2O5] = concentration of N2O5 1 = order with respect to N2O5 Rate and the order are obtained by experiments
7. Order The power of the concentrations is the order with respect to the reactant. E.g. a A+ b B-----> c C If the rate law: rate = k [A]1[B]2 The order of the reaction with respect to A is one (1). The order of the reaction with respect to B is two (2). Overall order of a chemical reaction is equal to the sum of all orders (3).
8. 1 [A]0 Graphical method 1 [A]t 1 [A]t
9. Rate Law Differential Rate Law Integral Rate rate = k [A]0 -D [A]/Dt= k ; ([A]0=1) [A]f-[A]0= -kt - d [A]/dt = k ; ([A]0=1 [A]f= -kt +[A]0 [A]f- [A]0= -kt rate= k [A]1-D[A]/D t = k [A] ln [A]t/[A]0= - kt d[A]/dt= - k [A] rate= k [A]2 -D [A]/Dt = k [A]21/ [A]f - 1/[A]0 = kt d[A]/dt= - k [A]2 1/ [A]f= kt - 1/[A]0 Differential and Integral Rate Law
10. Integral Law [A]f-[A]0= -kt ln[A]t/[A]0= -kt 1/[A]f = kt + 1/[A]0 Integral and Half-life forms t½Law t½ =  [A] o / 2k t½ = 0.693 / k t½ = 1 / k [A]o Zero order Firstorder Second order
11. 1) The reaction A ---> B + C is known to follow the rate law: rate = k [A]1 What are the differential, integral and half-life (t½) form of this rate law?
12. Comparing graphs This plot of ln[cis-platin] vs. time produces a straight line, suggesting that the reaction is first-order.
13. 2. Using graphical method, show that 2 N2O5 ---> 4 NO2 + O2, is a first order reaction.
14. Findingrate laws by Initial rates Method of initial rates The order for each reactant is found by: Changing the initial concentration of that reactant. Holding all other initial concentrations and conditions constant. Measuring the initial rates of reaction The change in rate is used to determine the order for thatspecific reactant. The process is repeated for each reactant.
15. Decomposition Reaction
16. Graphical Ways to get Order
17. Initial rate
18. How do get order of reactants E.g. a A+ b B-----> c C Hold [B] constant and change (double) [A] a A + b B -----> c C If the rate law: rate = k [A]x[B]y rate = k [A]1 k1 First order:1 x rate = k [2A]1 k1 = k 21[A]1 k1 rate1= k[A]1k1 rate1 = 1 rate2 = k 21[A]1k1 rate2= 21 (doubles) Second order:2 x rate = k [2A]1 k1 = k 22[A]2 k1 rate1 = k[A]2k1 rate1 = 1 rate2 = k22[A]2k1 rate2 = 22 (quadruples)
19. How do you find order? A + B -----> C rate = k [A]l[B]m; Hold concentration of other reactants constant If [A] doubled, rate doubled 1st order, [2A]1 = 2 1 x [A]1 , 2 1 = 2 b) If [A] doubled, rate quadrupled 2nd order, [2A]2 = 2 2 x [A]2 , 2 2 = 4 c) If [A] doubled, rate increased 8 times 3rd order, [2A]3 = 2 3 x [A]3 , 2 3 = 8
20. Rate data
21. 3. For the reaction: A ---> D, Find the order of [A] for each case. It was found in separate experiments that a) The rate doubled when [A] doubled b) The rate tripled when [A] tripled c) The rate quadrupled when [A] doubled d) The rate increased 8 times when [A] doubled
22. Units of the Rate Constant (k) 1 first order: k =───=s-1 s L second order k =─── mol s L2 third order k =─── mol2 s
23. 4. For the chemical reaction: A + B ----> C Using the following initial data to deduce: a) Order of each reactant b) Rate constant [A],mol/L [B],mol/L rate,mol/Ls _____________________________ 2.0 3.0 0.10 6.0 3.0 0.90 6.0 6.0 0.90
24. Overall order
25. Rate Constant E.g. a A+ b B-----> c C rate a [A]l[B]m rate = k [A]l[B]m; k = rate constant proportionality constant of the rate law Larger the k faster the reaction It is related inversely to t½
26. DeterminingK, Rate Constant
27. First Order Reactions and t½ A ----> B
28. Radio Activity and Nuclear Kinetics Nuclear reactions? Fusion Fission What kinetics fission follow?
29. Half-life t½
30. Nuclear Reactions : First order kinetics
31. t1/2 equation 0.693 = k t1/2 0.693 t1/2 = ---- k
32. Half-life -t1/2 The half-life and the rate constant are related. t1/2 = Half-life can be used to calculate the first order rate constant. For our N2O5 example, the reaction took 1900 seconds to react half way so: k = = = 3.65 x 10-4 s-1 0.693 k 0.693 t1/2 0.693 1900 s
33. 5. The rate constant for the first-order conversion of A to B is 2.22 hr-1. How much time will be required for the concentration of A to reach 75% of its original value?
34. 6) The half-life of a radioactive (follows first order rate law) isotope is 10 days. How many days would be required for the isotope to degrade to one eighth of its original radioactivity?
35. 7) The rate constant for the first order decomposition of SO2Cl2 (SO2Cl2SO2 +Cl2) at very high temperature is 1.37 × 10-3 min-1. If the initial concentration is 0.500 M, predict the concentration after five hours (300 min).