# 1 © 2006 Brooks/Cole - Thomson Chemical Kinetics Chapter 15 H 2 O 2 decomposition in an insect H 2 O 2 decomposition catalyzed by MnO 2.

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1 © 2006 Brooks/Cole - Thomson Chemical Kinetics Chapter 15 H 2 O 2 decomposition in an insect H 2 O 2 decomposition catalyzed by MnO 2

2 © 2006 Brooks/Cole - Thomson Interpreting Rate Laws Rate = k [A] m [B] n [C] p If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A Rate = k [A] 1 If [A] doubles, then rate goes up by factor of __ If [A] doubles, then rate goes up by factor of __ If m = 2, rxn. is 2nd order in A.If m = 2, rxn. is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by ________ If m = 0, rxn. is zero order.If m = 0, rxn. is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate ________

3 © 2006 Brooks/Cole - Thomson Concentration/Time Relations Integrating - ( [A] / time) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.

4 © 2006 Brooks/Cole - Thomson Properties of Reactions page 719

5 © 2006 Brooks/Cole - Thomson Concentration/Time Relations Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] Glucose If k = 0.21 hr -1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)?

6 © 2006 Brooks/Cole - Thomson Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law ln (0.100) = - 2.3 = - (0.21 hr -1 ) time time = 11 hours

7 © 2006 Brooks/Cole - Thomson Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Time (min)[N 2 O 5 ] 0 (M) ln [N 2 O 5 ] 0 01.000 1.00.705-0.35 2.00.497-0.70 5.00.173-1.75 Rate = k [N 2 O 5 ]

8 © 2006 Brooks/Cole - Thomson Using the Integrated Rate Law 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] Data of conc. vs. time plot do not fit straight line. Plot of ln [N 2 O 5 ] vs. time is a straight line!

9 © 2006 Brooks/Cole - Thomson Using the Integrated Rate Law All 1st order reactions have straight line plot for ln [A] vs. time. (2nd order gives straight line for plot of 1/[A] vs. time) Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = mx + b y = mx + b Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = mx + b y = mx + b

10 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful. Active Figure 15.9

11 © 2006 Brooks/Cole - Thomson Reaction is 1st order decomposition of H 2 O 2.Reaction is 1st order decomposition of H 2 O 2. Half-Life

12 © 2006 Brooks/Cole - Thomson Half-Life Reaction after 1 half-life.Reaction after 1 half-life. 1/2 of the reactant has been consumed and 1/2 remains.1/2 of the reactant has been consumed and 1/2 remains.

13 © 2006 Brooks/Cole - Thomson Half-Life After 2 half-lives 1/4 of the reactant remains.After 2 half-lives 1/4 of the reactant remains.

14 © 2006 Brooks/Cole - Thomson Half-Life A 3 half-lives 1/8 of the reactant remains.A 3 half-lives 1/8 of the reactant remains.

15 © 2006 Brooks/Cole - Thomson Half-Life After 4 half-lives 1/16 of the reactant remains.After 4 half-lives 1/16 of the reactant remains.

16 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Sugar is fermented in a 1st order process (using an enzyme as a catalyst). Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products sugar + enzyme --> products Rate of disappear of sugar = k[sugar] k = 3.3 x 10 -4 sec -1 What is the half-life of this reaction?

17 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution [A] / [A] 0 = fraction remaining when t = t 1/2 then fraction remaining = _________ Therefore, ln (1/2) = - k t 1/2 - 0.693 = - k t 1/2 t 1/2 = 0.693 / k t 1/2 = 0.693 / k So, for sugar, t 1/2 = 0.693 / k = 2100 sec = 35 min Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. What is the half- life of this reaction?

18 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution 2 hr and 20 min = 4 half-lives Half-life Time ElapsedMass Left 1st35 min2.50 g 2nd701.25 g 3rd1050.625 g 4th1400.313 g Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?

19 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Radioactive decay is a first order process. Tritium ---> electron + helium Tritium ---> electron + helium 3 H 0 -1 e 3 He 3 H 0 -1 e 3 He t 1/2 = 12.3 years If you have 1.50 mg of tritium, how much is left after 49.2 years?

20 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution ln [A] / [A] 0 = -kt [A] = ?[A] 0 = 1.50 mgt = 49.2 y Need k, so we calc k from: k = 0.693 / t 1/2 Obtain k = 0.0564 y -1 Now ln [A] / [A] 0 = -kt = - (0.0564 y -1 ) (49.2 y) = - 2.77 = - 2.77 Take antilog: [A] / [A] 0 = e -2.77 = 0.0627 0.0627 = fraction remaining Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

21 © 2006 Brooks/Cole - Thomson Half-Life Section 15.4 & Screen 15.8 Solution [A] / [A] 0 = 0.0627 0.0627 is the fraction remaining! Because [A] 0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> 0.750 mg after 1 half-life 1.50 mg ---> 0.750 mg after 1 half-life ---> 0.375 mg after 2 ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.188 mg after 3 ---> 0.094 mg after 4 ---> 0.094 mg after 4 Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

22 © 2006 Brooks/Cole - Thomson Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 238 U --> 234 Th + He4.5 x 10 9 y 14 C --> 14 N + beta5730 y 131 I --> 131 Xe + beta8.05 d Element 106 - seaborgium 263 Sg 0.9 s

23 © 2006 Brooks/Cole - Thomson Other rxn order half life equations For a zero order rxn For a 2 nd order rxn

24 © 2006 Brooks/Cole - Thomson MECHANISMS A Microscopic View of Reactions Sections 15.5 and 15.6 Mechanism: how reactants are converted to products at the molecular level. RATE LAW ----> MECHANISM experiment ---->theory

25 © 2006 Brooks/Cole - Thomson MECHANISMS & Activation Energy Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene]

26 © 2006 Brooks/Cole - Thomson MECHANISMS Activation energy barrier CisTrans Transition state

27 © 2006 Brooks/Cole - Thomson Energy involved in conversion of trans to cis butene trans cis energy Activated Complex +262 kJ -266 kJ MECHANISMS See Figure 15.14 4 kJ/mol

28 © 2006 Brooks/Cole - Thomson Mechanisms Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product.Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product. ACTIVATION ENERGY, E a = energy reqd to form activated complex. Here E a = 262 kJ/mol

29 © 2006 Brooks/Cole - Thomson

30 © 2006 Brooks/Cole - Thomson Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis ---> trans is EXOTHERMIC This is the connection between thermo- dynamics and kinetics. MECHANISMS

31 © 2006 Brooks/Cole - Thomson Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, E a. Reaction coordinate diagram

32 © 2006 Brooks/Cole - Thomson More About Activation Energy Arrhenius equation Arrhenius equation Rate constant Temp (K) 8.31 x 10 -3 kJ/Kmol Activation energy Frequency factor Frequency factor related to frequency of collisions with correct geometry. Plot ln k vs. 1/T ---> straight line. slope = -E a /R

33 © 2006 Brooks/Cole - Thomson In addition to graphically, E a can be calculated algebraically If you know k at two different temperatures, we can calculate it

34 © 2006 Brooks/Cole - Thomson Effect of Temperature Reactions generally occur slower at lower T.Reactions generally occur slower at lower T. Iodine clock reaction, Screen 15.11, and book page 705. H 2 O 2 + 2 I - + 2 H + --> 2 H 2 O + I 2 Room temperature In ice at 0 o C

35 © 2006 Brooks/Cole - Thomson Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. In general, differences in activation energy cause reactions to vary from fast to slow.

36 © 2006 Brooks/Cole - Thomson CATALYSIS Catalysis and activation energy Uncatalyzed reaction Catalyzed reaction MnO 2 catalyzes decomposition of H 2 O 2 2 H 2 O 2 ---> 2 H 2 O + O 2

37 © 2006 Brooks/Cole - Thomson Iodine-Catalyzed Isomerization of cis-2-Butene Figure 15.16

38 © 2006 Brooks/Cole - Thomson Iodine-Catalyzed Isomerization of cis-2-Butene

39 © 2006 Brooks/Cole - Thomson

40 © 2006 Brooks/Cole - Thomson Mechanisms 1.Why is trans-butene cis-butene reaction observed to be 1st order? As [trans] doubles, number of molecules with enough E also doubles. As [trans] doubles, number of molecules with enough E also doubles. 2.Why is the trans cis reaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T. Fraction of molecules with sufficient activation energy increases with T.

41 © 2006 Brooks/Cole - Thomson More on Mechanisms Reaction of trans-butene --> cis-butene is UNIMOLECULAR - only one reactant is involved. BIMOLECULAR two different molecules must collide --> products A bimolecular reaction Exo- or endothermic?

42 © 2006 Brooks/Cole - Thomson Collision Theory Reactions require (a) activation energy and (b) correct geometry. O 3 (g) + NO(g) ---> O 2 (g) + NO 2 (g) O 3 (g) + NO(g) ---> O 2 (g) + NO 2 (g) 2. Activation energy and geometry 1. Activation energy

43 © 2006 Brooks/Cole - Thomson Mechanisms O 3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction.

44 © 2006 Brooks/Cole - Thomson Mechanisms Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ]NOTE 1.Rate law comes from experiment 2.Order and stoichiometric coefficients not necessarily the same! 3.Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

45 © 2006 Brooks/Cole - Thomson Mechanisms Proposed Mechanism Step 1 slowHOOH + I - --> HOI + OH - Step 2 fastHOI + I - --> I 2 + OH - Step 3 fast2 OH - + 2 H + --> 2 H 2 O Rate of the reaction controlled by slow step Rate of the reaction controlled by slow step RATE DETERMINING STEP, rds. RATE DETERMINING STEP, rds. Rate can be no faster than rds! Proposed Mechanism Step 1 slowHOOH + I - --> HOI + OH - Step 2 fastHOI + I - --> I 2 + OH - Step 3 fast2 OH - + 2 H + --> 2 H 2 O Rate of the reaction controlled by slow step Rate of the reaction controlled by slow step RATE DETERMINING STEP, rds. RATE DETERMINING STEP, rds. Rate can be no faster than rds! Most rxns. involve a sequence of elementary steps. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ]

46 © 2006 Brooks/Cole - Thomson Mechanisms Elementary Step 1 is bimolecular and involves I - and HOOH. Therefore, this predicts the rate law should be Rate [I - ] [H 2 O 2 ] as observed!! The species HOI and OH - are reaction intermediates. Elementary Step 1 is bimolecular and involves I - and HOOH. Therefore, this predicts the rate law should be Rate [I - ] [H 2 O 2 ] as observed!! The species HOI and OH - are reaction intermediates. 2 I - + H 2 O 2 + 2 H + ---> I 2 + 2 H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ] Step 1 slowHOOH + I - --> HOI + OH - Step 2 fastHOI + I - --> I 2 + OH - Step 3 fast2 OH - + 2 H + --> 2 H 2 O

47 © 2006 Brooks/Cole - Thomson Rate Laws and Mechanisms NO 2 + CO reaction: Rate = k[NO 2 ] 2 Single step Two possible mechanisms Two steps: step 1 Two steps: step 2

48 © 2006 Brooks/Cole - Thomson Ozone Decomposition over Antarctica 2 O 3 (g) ---> 3 O 2 (g)

49 © 2006 Brooks/Cole - Thomson Ozone Decomposition Mechanism Proposed mechanism Step 1: fast, equilibrium O 3 (g) Æ O 2 (g) + O (g) Step 2: slowO 3 (g) + O (g) ---> 2 O 2 (g) 2 O 3 (g) ---> 3 O 2 (g)

50 © 2006 Brooks/Cole - Thomson CATALYSISCATALYSIS Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier. Dr. James Cusumano, Catalytica Inc. What is a catalyst? Catalysts and society Catalysts and the environment

51 © 2006 Brooks/Cole - Thomson CATALYSIS In auto exhaust systems Pt, NiO 2 CO + O 2 ---> 2 CO 2 2 NO ---> N 2 + O 2

52 © 2006 Brooks/Cole - Thomson CATALYSIS 2.Polymers: H 2 C=CH 2 ---> polyethylene 3.Acetic acid: CH 3 OH + CO --> CH 3 CO 2 H 4. Enzymes biological catalysts 2.Polymers: H 2 C=CH 2 ---> polyethylene 3.Acetic acid: CH 3 OH + CO --> CH 3 CO 2 H 4. Enzymes biological catalysts

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