Chemical Kinetics: A Preview

1. Chemical Kinetics: A Preview • Chemical kinetics is the study of: • the rates of chemical reactions • factors that affect these rates • the mechanisms by which reactions occur • Reaction rates vary greatly – some are very fast (burning, precipitation) and some are very slow (rusting, disintegration of a plastic bottle in sunlight).

2. Reaction Rates Chemical Kinetics: The area of chemistry concerned with reaction rates and the sequence of steps by which reactions occur. Reaction Rate: Either the increase in the concentration of a product per unit time or the decrease in the concentration of a reactant per unit time.

3. 2N2O5(g) 4NO2(g) + O2(g) Reaction Rates decrease increase

4. 2N2O5(g) 4NO2(g) + O2(g) Reaction Rates

5. M -(0.0101 M - 0.0120 M) s 2N2O5(g) 4NO2(g) + O2(g) (400 s - 300 s) D[N2O5] Dt Reaction Rates Rate of decomposition of N2O5: = = 1.9 x 10-5

6. D[NO2] D[O2] D[N2O5] Dt Dt Dt a A + b B 2N2O5(g) d D + e E 4NO2(g) + O2(g) D[D] D[B] D[E] D[A] Dt Dt Dt Dt 1 1 1 1 1 1 d 2 e 4 b a Reaction Rates General rate of reaction: rate = - = = rate = - = - = =

7. 2N2O5(g) 4NO2(g) + O2(g) Reaction Rates

8. rate = k[A]m[B]n aA + bB products a rate [A]m[B]n Rate Laws and Reaction Order Rate Law: An equation that shows the dependence of the reaction rate on the concentration of each reactant. k is the rate constant

9. Rate Laws and Reaction Order The values of the exponents in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

10. 2NO(g) + O2(g) 2NO2(g) rate = k[NO]m [O2]n Experimental Determination of a Rate Law Compare the initial rates to the changes in initial concentrations.

11. 2NO(g) + O2(g) 2NO2(g) rate = k[NO]2 [O2]n Experimental Determination of a Rate Law The concentration of NO doubles, the concentration of O2 remains constant, and the rate quadruples. 2m = 4 m = 2

12. 2NO(g) + O2(g) 2NO2(g) rate = k[NO]2 [O2] Experimental Determination of a Rate Law The concentration of O2doubles, the concentration of NO remains constant, and the rate doubles. 2n = 2 n = 1

13. 2NO(g) + O2(g) 2NO2(g) rate = k[NO]2 [O2] Experimental Determination of a Rate Law Reaction Order With Respect to a Reactant • NO: second-order • O2: first-order Overall Reaction Order • 2 + 1 = 3 (third-order)

14. 2NO(g) + O2(g) 2NO2(g) rate = k[NO]2 [O2] Experimental Determination of a Rate Law

15. Moving along…rate laws/integrated • 1st determined the ‘order’ of a reaction (experimentally); loss of reactant • derived rate law (determined exponents) • Solved for rate constant • REMEMBER…make sure you understand the diff. btw • rate constant • rate LAW (generic form AND the ‘solved’ form) • initial rate • Ultimately, we want to use this info to determine [ ]’s in the future. We do this with ‘integrated rate laws’.

16. A product(s) [A]t x ln = -kt ln = ln(x) - ln(y) [A]0 y D[A] - = k[A] Dt Integrated Rate Law for a First-Order Reaction rate = k[A] Calculus can be used to derive an integrated rate law. [A]t concentration of A at time t [A]0 initial concentration of A Using: ln[A]t = -kt + ln[A]0 y = mx + b

17. Integrated Rate Law for a First-Order Reaction ln[A]t = -kt + ln[A]0 y = mx + b A plot of ln[A] versus time gives a straight-line fit and the slope will be -k.

18. Integrated Rate Law for a First-Order Reaction ln[A]t = -kt + ln[A]0 This is a plot of [A] versus time. The best-fit is a curve and not a line.

19. Integrated Rate Law for a First-Order Reaction ln[A]t = -kt + ln[A]0

20. 2N2O5(g) 4NO2(g) + O2(g) -5.099 - (-3.912) (700 - 0) s 1 1 s s Integrated Rate Law for a First-Order Reaction rate = k[N2O5] Calculate the slope: Slope = -k = -0.0017 k = 0.00170

21. A product(s) 1 = -kt1/2 ln [A]t 2 ln = -kt [A]0 [A]0 = [A] 0.693 t1/2 2 t1/2 = k Half-Life for a First-Order Reaction Half-Life: The time required for the reactant concentration to drop to one-half of its initial value. rate = k[A] t = t1/2 or

22. 0.693 t1/2 = k Half-Life for a First-Order Reaction For a first-order reaction, the half-life is independent of the initial concentration. Each successive half-life is an equal period of time.

23. Nuclear Reactions and Radioactivity Alpha (a ) Radiation An alpha particle is a helium-4 nucleus (2 protons and 2 neutrons). Alpha particle, a

24. Nuclear Reactions and Radioactivity In nuclear equations, nuclear particles must balance. (not the elements…as you see below) 238 nucleons 92 protons 4 nucleons 2 protons 234 nucleons 90 protons +

25. Nuclear Reactions and Radioactivity Beta (b ) Radiation A beta particle is an electron. Beta particle, b -

26. Nuclear Reactions and Radioactivity Gamma (g ) Radiation A gamma particle is a high-energy photon Positron Emission A positron has the same mass as an electron but an opposite charge. It can be thought of as a “positive electron.” Positron, b +

27. Nuclear Reactions and Radioactivity Electron Capture A process in which the nucleus captures an inner-shell electron, thereby converting a proton to a neutron.

28. Nuclear Reactions and Radioactivity

29. N ln = -kt N0 Radioactive Decay Rates— This is why we’re looking at nuclear boom-boom stuff Radioactive decay is a first-order KINETIC process. Decay = k x N k is the decay constant. N0 is the number of radioactive nuclei initially present in a sample. N is the number of radioactive nuclei remaining at time t.

30. ln 2 ln 2 t1/2 = k = t1/2 k Radioactive Decay Rates Radioactive decay is a first-order process. Half-life: or

31. So we did a couple of problems • Yesterday…a few problems. Do others on your own! • Today…what’s on tap. The last problem from the ‘worksheet’…and then… • Carbon 13, 12 and 14 • A nuclear schematic…why? Cuz I think it’s cool. • Second and Zero order processes (and a sample problem) • Questions YOU may have about the quiz Friday. • Material for said quiz ends with stuff from today.

32. Radioactive Decay Rates

33. Nuclear Fission and Fusion Nuclear Fission

34. 142 235 91 1 1 3 + + + Ba n n Kr U 92 56 36 0 0 Nuclear Fission and Fusion Nuclear Fission Chain Reaction: A self-sustaining reaction whose product initiates further reaction.

36. A Zero-Order Reaction rate = k[A]0 = k Rate is independent of initial concentration

37. Second-Order Reactions • A reaction that is second order in a reactant has a rate law in which the exponent for that reactant is 2. Rate = k[A]2 • The integrated rate law has the form: What do we plot vs. time to get a straight line? 1 1 –––– = kt + –––– [A]t [A]0 • The half-life of a second-order reaction depends on the initial concentration as well as on the rate constant k: 1 t½ = ––––– k[A]0

38. Example The second-order decomposition of HI(g) at 700 K is represented in Figure 13.9. HI(g)  ½ H2(g) + ½ I2(g) Rate = k[HI]2 What are the: (a) rate constant and (b) half-life of the decomposition of 1.00 M HI(g) at 700 K?

39. Another Summary of Kinetic Data Your sample problems from Tuesday’s class has another…this has zero order

40. 2N2O5(g) 4NO2(g) + O2(g) The Arrhenius Equation Typically, as the temperature increases, the rate of reaction increases. rate = k[N2O5] The rate constant is dependent on temperature.

41. Theories of Chemical Kinetics: Collision Theory • Before atoms, molecules, or ions can react, they must first collide. • An effective collision between two molecules puts enough energy into key bonds to break them. • The activation energy (Ea) is the minimum energy that must be supplied by collisions for a reaction to occur. • A certain fraction of all molecules in a sample will have the necessary activation energy to react; that fraction increases with increasing temperature. • The spatial orientationsof the colliding species may also determine whether a collision is effective.

42. Importance of Orientation H + H → H2 CH3Br + I-→ CH3I + Br- Effective collision; “backside attack” Ineffective collision; BOINK! … rxn occurs…red hydrogen doesn’t really care…like it’s wearing a big KICK ME sign!!

43. Effect of Temperature onthe Rates of Reactions • In 1889, Svante Arrhenius proposed the following expression for the effect of temperature on the rate constant, k: k = Ae–Ea/RT • The constant A, called the frequency factor, is an expression of collision frequency and orientation; it represents the number of collisions per unit time that are capable of leading to reaction. • The term e–Ea/RT represents the fraction of molecular collisions sufficiently energetic to produce a reaction. • This is where the rate ‘constant’ is variable.

44. Distribution of Kinetic Energies At higher temperature (red), more molecules have the necessary activation energy.

45. Transition State Theory • The configuration of the atoms of the colliding species at the time of the collision is called the transition state. • The transitory species having this configuration is called the activated complex. • A reaction profile shows potential energy plotted as a function of a parameter called the progress of the reaction…recall the Halloween analogy…pixie stix • Reactant molecules must have enough energy to surmount the energy “hill” separating products from reactants.

46. A Reaction Profile CO(g) + NO2(g)  CO2(g) + NO(g)

47. Think about that for a sec… • Recall that it takes 3-4 experiments to determine a rate law at a constant temp, THEN it takes 3-4 temps to determine the activation energy!! Can be quite time consuming!! • The Arrhenius equation takes several forms (take ln of both sides—can make a plot of ln k as a function of Ea/R (NOTE: R = 8.314 J / mol K—not ideal gas!) • First order decomposition of DTBP results in a half-life of 17.5 h at 125 °C and 1.67 h at 167 °C. Find Ea.

48. a k = Ae-E /RT The Arrhenius Equation k rate constant A collision frequency factor Ea activation energy R gas constant T temperature (K)

49. a ln(k) = ln(A) + ln(e-E /RT) Ea ln(k) = ln(A) - RT 1 ln(k) = + ln(A) T -Ea R Using the Arrhenius Equation rearrange the equation y = mx + b

50. -Ea R 1 T 1 ln(k) = + ln(A) T -Ea Slope = R Using the Arrhenius Equation Plot ln(k) versus