Ch #16: Kinetics

1. Ch #16: Kinetics • Kinetics: Rates and Mechanisms of Chemical Reactions

2. Chemical kinetics: • Study of reaction rates, changes in concentrations of reactants or products as a function of time.

3. Factors that influence reaction rate: 1.Concentration - molecules must collide to react; Rate α collision frequency α conc 2.Physical state - molecules must mix to collide; Smaller the particle size, greater the surface area and more the collisions. 3.Temperature - molecules must collide with enough energy to react; Rate α collision frequency α conc.

4. The effect of surface area on reaction rate. Figure 16.3

5. Collision energy and reaction rate. Figure 16.4

6. Factors that influence reaction rate: 4.The use of a catalyst.

7. Expressing the Reaction Rate • reaction rate - changes in the concentrations of reactants or products per unit time • reactant concentrations decrease while product concentrations increase • For a reaction A → B • Rate of reaction = -∆[A]/∆t • [A]= conc of A in mol/L • ∆=change

8. Expressing the Reaction Rate • Units= moles per liter per second. • Mol L-1s-1 or mol/ L.s • Change I product conc is positive so the rate is Rate of reaction = ∆[B]/∆t • Rate decreases during course of reaction. • Instantaneous rate: Rate at a particular instant: considering closer values.

9. Expressing the Reaction Rate • Use initial rate as soon as the reactants are present( no products at this time) • For a reaction aA + bB →cC+ dD, • Rate =-1/a∆[A]/∆t=-1/b∆[B]/∆t==1/c∆[C]/∆t==1/d∆[A]/∆t

10.  (conc A) - t Table 16.1 Concentration of O3 at Various Time in its Reaction with C2H4 at 303K Concentration of O3 (mol/L) Time (s) 0.0 3.20x10-5 10.0 2.42x10-5 20.0 1.95x10-5 30.0 1.63x10-5 40.0 1.40x10-5 50.0 1.23x10-5 60.0 1.10x10-5

11.  [C2H4] - t  [O3] - t Figure 16.5 The concentrations of O3 vs. time during its reaction with C2H4 rate = =

12. Plots of [C2H4] and [O2] vs. time. Figure 16.6 Tools of the Laboratory

13. Problem 1 • P.1.2H2 (g) +O2 (g)→ 2H2O (g) Write the rate equation in terms of reactants and products. If [O2] is decreasing at 0.23 mol/L .s at what rate is [H2O] increasing?

14. The Rate Law and its components: • Rate = k[A]m [B]n • K= rate constant(does not change as reaction proceeds) • M and n are reaction orders. • Coefficients are not related to reaction orders. • Rate constant and orders can only be found by experimental data.

15. Determining the Initial Rate: • By performing expt and collecting data.

16. Reaction order: • Rate=k[A]- 1st order • Rate = k[A]2- 2nd order • Rate = k[A]0- Zero order OR Rate=k(1)=k (not dependent on conc of A)

17. Problem 2 • P.2.For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. • 2NO(g) + O2(g) → 2NO2(g); rate = k[NO]2[O2] • CH3CHO(g) → CH4(g) + CO(g); rate = k[CH3CHO]3/2

18. Problem 2 • H2O2(aq) + 3I-(aq) + 2H+(aq) → I3-(aq) + 2H2O(l); rate = k[H2O2][I-]

19. Determining Reaction Orders: • P.3.NO2(g)+ CO (g) → NO (g) + CO2(g) rate=k[NO2]m [CO]n

20. PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these is rate = k[NO2]m[CO]n Experiment Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L) 1 0.10 0.0050 0.10 2 0.10 0.080 0.40 3 0.0050 0.10 0.20 Sample Problem 16.3 Determining Reaction Order from Initial Rate Data Use the following data to determine the individual and overall reaction orders. PLAN: Solve for each reactant using the general rate law using the method described previously. SOLUTION: rate = k [NO2]m[CO]n First, choose two experiments in which [CO] remains constant and the [NO2] varies.

21. n m rate 2 k [NO2]m2[CO]n2 [CO]3 [NO2]2 = = rate 1 k [NO2]m1 [CO]n1 [CO]1 [NO2]1 m 0.080 0.40 = 0.0050 0.10 rate 3 k [NO2]m3[CO]n3 = = rate 1 k [NO2]m1 [CO]n1 n 0.0050 0.20 = 0.0050 0.10 Sample Problem 16.3 Determining Reaction Order from Initial Rate Data continued The reaction is 2nd order in NO2. ; 16 = 4m and m = 2 The reaction is zero order in CO. ; 1 = 2n and n = 0 rate = k [NO2]2[CO]0 = k [NO2]2

22. Determining the Rate constant: • Calculated from collected data. • Units of Rate constant depend on order.

23. Table 16.4 An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions Zero Order First Order Second Order rate = k Rate law rate = k [A] rate = k [A]2 mol/L*s Units for k 1/s L/mol*s [A]t = k t + [A]0 ln[A]t = -k t + ln[A]0 1/[A]t = k t + 1/[A]0 Integrated rate law in straight-line form [A]t vs. t ln[A]t vs. t 1/[A]t = t Plot for straight line k, 1/[A]0 Slope, y-intercept k, [A]0 -k, ln[A]0 Half-life [A]0/2k ln 2/k 1/k [A]0

24. Integrated rate laws and reaction order Figure 16.7 1/[A]t = kt + 1/[A]0 [A]t = -kt + [A]0 ln[A]t = -kt + ln[A]0

25. Problem 5 • P.5At 1000 o C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87s-1, to two molecules of ethylene (C2H4). • a) If the initial C4H8 concentration is 2.00M, what is the concentration after 0.010 s? • (b) What fraction of C4H8 has decomposed in this time?

26. Rules for orders through graphs • If a straight line is produced with the foll: • Ln[reactant] vs time-1st • 1/[reactant] vs time-2nd • [reactant]vs time -0 order.

27. Half Life • Half Life: Time required for the reactant concentration to reach half of its initial value.

28. for a first-order process ln 2 0.693 t1/2 = = k k A plot of [N2O5] vs. time for three half-lives. Figure 16.9

29. Problem 6 • P.6 Cyclopropane is the smallest cyclic hydrocarbon. Because its 600 bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 10000C via a first-order reaction: • The rate constant is 9.2s-1; (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value?

30. Effect of Temperature: • Temperature affects rate by affecting rate constant. • Arrhenius equ: k=Ae-Ea/RT • K=rate const, e=base of natural logs, T=absolute temp, R= Universal gas const, Ea=activation energy (minimum energy that molecules must have to react)

31. Effect of Temperature: • As T increases, k increases, rate increases. • Ln k =ln A- Ea/R(1/T) • A plot of ln k vs 1/T gives a straight line • Slope=-Ea/R • Y-int= ln A • Ln K2/k1=-Ea/R(1/T2-1/T1)

32. k2 Ea 1 1 ln = - - k1 RT T2 T1 The Effect of Temperature on Reaction Rate The Arrhenius Equation where k is the kinetic rate constant at T Ea is the activation energy R is the energy gas constant T is the Kelvin temperature ln k = ln A - Ea/RT A is the collision frequency factor

33. Graphical determination of the activation energy Figure 16.11 ln k = -Ea/R (1/T) + ln A

34. Problem 7 P.7.The decomposition of hydrogen iodide 2HI(g) → H2(g) + I2(g) • has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea.

35. Determine slope of tangent at t0 for each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Substitute initial rates, orders, and concentrations into general rate law: rate = k [A]m[B]n Find k at varied T Find k at varied T Rearrange to linear form and graph Use direct, ln or inverse plot to find order Figure 16.12 Information sequence to determine the kinetic parameters of a reaction. Series of plots of concentra-tion vs. time Initial rates Reaction orders Rate constant (k) and actual rate law Activation energy, Ea Rate constant and reaction order Integrated rate law (half-life, t1/2) Plots of concentration vs. time

36. Effects of concentration and temperature: • Collision Theory: Reactant particles must collide with each other. • Concs are multiplied in rate law. • Ea: Energy required to activate the molecules into a state from which reactant bonds can change into product bonds. • Only those collisions with enough energy to exceed Ea can lead to reaction.

37. A A A A A A A A B B B B B B B Figure 16.13 The dependence of possible collisions on the product of reactant concentrations. 4 collisions Add another molecule of A 6 collisions Add another molecule of B

38. Effects of concentration and temperature: • Rise in temp enlarges the fraction of collisions with enough energy to exceed Ea. • F= e-Ea/RT e-base of natural log, T=temp, R=gas const. • In exothermic reaction: Eaf > Ear • In endothermic reaction: Eaf < Ear • : k=Ae-Ea/RT A is the frequency factor, A=pZ, Z=collision frequency and p=orientation probability factor.

39. Figure 16.14 The effect of temperature on the distribution of collision energies

40. ACTIVATEDSTATE Ea (forward) Ea (reverse) Collision Energy Collision Energy REACTANTS PRODUCTS Energy-level diagram for a reaction Figure 16.15 The forward reaction is exothermic because the reactants have more energy than the products.

41. NO + NO3 2 NO2 Figure 16.17 The importance of molecular orientation to an effective collision. A is the frequency factor Z is the collision frequency p is the orientation probability factor A = pZ where

42. Transition state theory: • Transition state/activated complex: neither reactant nor product present, a transitional species with partial bonds is present. • Ea is the energy required to stretch and deform bonds in order to reach transition state. • Reaction energy diagram : TB. Pg.699

43. CH3Br + OH- CH3OH + Br - Figure 16.18 Nature of the transition state in the reaction between CH3Br and OH-. transition state or activated complex

44. Reaction energy diagram for the reaction of CH3Br and OH-. Figure 16.19

45. Figure 16.20 Reaction energy diagrams and possible transition states.

46. Molecularity of a reaction: • Elementary steps make up the reaction mechanism. • An elementary step is not made up of simpler steps. • Molecularity means the number of reactant particles involved in the step.

47. Molecularity of a reaction: • In an elementary step equation coeff= order. • Reaction order= molecularity • A→ product Unimolecular Rate = k[A] • 2A→ product Bimolecular Rate = k[A]2 • A + B→ product Bimolecular Rate = k[A][B] • 2A + B→ product Termolecular Rate =k [A]2[B]

48. A product 2A product A + B product 2A + B product REACTION MECHANISMS Table 16.6 Rate Laws for General Elementary Steps Elementary Step Molecularity Rate Law Unimolecular Rate = k [A] Bimolecular Rate = k [A]2 Rate = k [A][B] Bimolecular Termolecular Rate = k [A]2[B]

49. Problem 8 • P.8. The following two reactions are proposed as elementary steps in the mechanism of an overall reaction:NO2Cl(g) → NO2(g) + Cl(g)NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g)(a) Write the overall balanced equation. • NO2Cl(g) → NO2(g) + Cl(g) • NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g)

50. Problem 8 (a) Write the overall balanced equation. • Determine the molecularity of each step. • Write the rate law for each step