1 KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. Chemical Kinetics Chapter 15

2 Reaction Mechanisms The sequence of events at the molecular level that control the speed and outcome of a reaction.

3 change in concentration of a reactant or product with timechange in concentration of a reactant or product with time initial rateinitial rate average rate average rate instantaneous rateinstantaneous rate Reaction Rates

4 Concentration/Time Relations Need to know what conc. of reactant is as function of time. Consider FIRST ORDER REACTIONS For 1st order reactions, the rate law is - ( [A] / time) = k [A] For 1st order reactions, the rate law is - (  [A] /  time) = k [A] Need to know what conc. of reactant is as function of time. Consider FIRST ORDER REACTIONS For 1st order reactions, the rate law is - ( [A] / time) = k [A] For 1st order reactions, the rate law is - (  [A] /  time) = k [A]

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

6 Half-Life Section 15.4 and 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.

7 Half-LifeHalf-Life Reaction after 654 min, 1 half-life.Reaction after 654 min, 1 half-life. 1/2 of the reactant remains.1/2 of the reactant remains.

8 Half-LifeHalf-Life Reaction after 1306 min, or 2 half-lives.Reaction after 1306 min, or 2 half-lives. 1/4 of the reactant remains.1/4 of the reactant remains.

9 Half-LifeHalf-Life Reaction after 3 half-lives, or 1962 min.Reaction after 3 half-lives, or 1962 min. 1/8 of the reactant remains.1/8 of the reactant remains.

10 Half-Life Section 15.4 and Screen 15.8 Solution [A] / [A] 0 = 1/2 when t = t 1/2 Therefore, ln (1/2) = - k t 1/ = - k t 1/2 t 1/2 = / k t 1/2 = / k So, for sugar, t 1/2 = / k = 2100 sec = 35 min Rate = k[sugar] and k = 3.3 x sec -1. What is the half- life of this reaction?

11 Half-Life Section 15.4 and Screen 15.8 Solution 2 hr and 20 min = 4 half-lives Half-life Time ElapsedMass Left 1st35 min5.00 g 2nd g 3rd g 4th g Rate = k[sugar] and k = 3.3 x sec -1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min?

12 Half-Life Section 15.4 and Screen 15.8 Solution ln [A] / [A] 0 = -kt [A] = ?[A] 0 = 1.50 mgt = 49.2 mg Need k, so we calc k from: k = / t 1/2 Obtain k = y -1 Now ln [A] / [A] 0 = -kt = - ( y -1 ) (49.2 y) = = Take antilog: [A] / [A] 0 = e = is the fraction remaining! Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

13 Concentrations and physical state of reactants and productsConcentrations and physical state of reactants and products TemperatureTemperature CatalystsCatalysts Factors Affecting Rates Section 15.2

14 Collision Theory Reactions require (a) activation energy and (b) correct geometry. O 3 (g) + NO(g) ---> O 2 (g) + NO 2 (g) Reactions require (a) activation energy and (b) correct geometry. O 3 (g) + NO(g) ---> O 2 (g) + NO 2 (g) O 3 + NO reaction occurs in a single ELEMENTARY step. single ELEMENTARY step.

15 Collision Theory explains effects Of Conc. & Temp on Rates! Molecules must collide Molecules must collide with enough energy Molecules must collide with the right orientation

16 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) 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)

17 Concentrations and Rates To postulate a reaction mechanism, we study reaction rate andreaction rate and its concentration dependenceits concentration dependence To postulate a reaction mechanism, we study reaction rate andreaction rate and its concentration dependenceits concentration dependence

18 Arrhenius Equation k = A e –Ea/Rt k = A e –Ea/Rt A = frequency of collisions with correct geometry. e –Ea/Rt e –Ea/Rt = fraction of molecules with mimimum energy for the reaction

19 Arrhenius Equation k = A e –Ea/RT ln k = ln A - (Ea/RT) ln k = ln A - (Ea/RT) ln k = ln A - Ea/R ( 1/T) ln k = ln A - Ea/R ( 1/T)

20 Arrhenius Equation As Temperature increases, the fraction of molecules with sufficient activation energy increases. Temp (K)e -Ea/Rt x x x 10 -4

21 MECHANISMS Sections 15.5 and 15.6 How are reactants converted to products at the molecular level? RATE LAW ----> MECHANISM experiment ---->theory How are reactants converted to products at the molecular level? RATE LAW ----> MECHANISM experiment ---->theory

22 MECHANISMSMECHANISMS For example Rate = k [trans-2-butene] Conversion requires twisting around the C=C bond. For example Rate = k [trans-2-butene] Conversion requires twisting around the C=C bond.

23 Energy involved in conversion of trans to cis butene MECHANISMSMECHANISMS See Figure 15.15

24 MECHANISMSMECHANISMS TRANSITION STATE ACTIVATION ENERGY, E a = energy req’d to form activated complex. Here E a = 233 kJ/mol TRANSITION STATE ACTIVATION ENERGY, E a = energy req’d to form activated complex. Here E a = 233 kJ/mol

25 Activation Energy Molecules are moving…..but how many of them have enough Energy to go to product? What does increasing T do? A flask full of trans-butene is stable because only a tiny fraction of trans molecules have enough energy to convert to cis. In general, differences in activation energy are the reason reactions vary from fast to slow. Molecules are moving…..but how many of them have enough Energy to go to product? What does increasing T do? A flask full of trans-butene is stable because only a tiny fraction of trans molecules have enough energy to convert to cis. In general, differences in activation energy are the reason reactions vary from fast to slow.

26 MECHANISMSMECHANISMS 1.Why is 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 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. 1.Why is 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 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.

27 Reaction of trans --> cis is UNIMOLECULAR- only one reactant is involved. BIMOLECULAR — two different molecules must collide --> products Reaction of trans --> cis is UNIMOLECULAR- only one reactant is involved. BIMOLECULAR — two different molecules must collide --> products MECHANISMSMECHANISMS

28 MECHANISMS BIMOLECULAR — two different molecules must collide --> products A bimolecular reaction Exo- or endothermic?

29 MECHANISMSMECHANISMS Most reactions involve a sequence of elementary steps. Adding elementary steps gives NET reaction. Most reactions involve a sequence of elementary steps. Adding elementary steps gives NET reaction.

30

31

32 Relationship to Reaction order Molecularity of an Elementary Step and its order are the same. Not necessarily true for overall reaction, just for elementary steps!

33 Rate Equations again A  unimolecularRate = k [A] A + B  BimolecularRate = k [A][B] A + A  BimolecularRate = k [A] 2 2 A + B  TermolecularRate = k [A] 2 [B]

34 MECHANISMSMECHANISMS 2 I - + H 2 O H + ---> I H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ] Step 1 —HOOH + I - --> HOI + OH - Step 2 —HOI + I - --> I 2 + OH - Step 3 —2 OH H + --> 2 H 2 O 2 I - + H 2 O H + ---> I H 2 O Rate = k [I - ] [H 2 O 2 ] Rate = k [I - ] [H 2 O 2 ] Step 1 —HOOH + I - --> HOI + OH - Step 2 —HOI + I - --> I 2 + OH - Step 3 —2 OH H + --> 2 H 2 O

35 MECHANISMSMECHANISMS 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. 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 H + ---> I 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 H + --> 2 H 2 O

36 Exercise NO  N 2 O 2 N 2 O 2 + H 2  N 2 O + H 2 O N 2 O + H 2  N 2 + H 2 O Molecularity? Rate Eqns? Sum of Steps?

37 Rate of the reaction controlled by slow step RATE DETERMINING STEP Rate can be no faster than RDS!

38 CATALYSISCATALYSIS Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier.

39 Catalysts in Industry Petroleum refining Industrial production of chemicals, pharmaceuticals Environmental controls Heterogeneous vs. Homogeneous

40 CATALYSISCATALYSIS In auto exhaust systems — Pt, NiO 2 CO + O 2 ---> 2 CO 2 2 NO ---> N 2 + O 2 In auto exhaust systems — Pt, NiO 2 CO + O 2 ---> 2 CO 2 2 NO ---> N 2 + O 2

41 CATALYSISCATALYSIS 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

42 CATALYSISCATALYSIS 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

43 MnO 2 catalyzes decomposition of H 2 O 2 Figure 15.18

44 Iodine-Catalyzed Isomerization of cis-2-Butene Figure 15.19

45

46 Cis-2-butene  trans-2-butene 1) I 2  2 I 2) I + cis-2-butene  I-cis-2-butene 3) I-cis-2-butene  I-trans-2-butene 4) I-trans-2-butene  I + trans-2-butene 5) I + I  I 2 Rate = k [cis-2-butene][I 2 ] 1/2 One I 2 gets broken, one I used, but regenerated. In the end, the two I can recombine. No net consumption of I 2 !