Kinetics (the study of reaction rates) Part 1

1. Kinetics (the study of reaction rates)Part 1 Ch. 14 in Text Images from: www.wiley.com/www.saskschools.ca/ www.fife-education.org.uk/ en.wikipedia.org/petersondavis.wordpress.com/ www.fauske.com/tehparadox.com/ www.arbdownload.com

2. What you (maybe) already know about kineticsA) Definitions • What is a reaction rate? • What is a reaction mechanism?

3. B) Kinetics Activity • How did this activity demonstrate reaction rate? • How did this activity demonstrate a mechanism? • Concentration vs. Time Plots

4. C) Factors Affecting Rxn Rate

5. D) Potential Energy Diagrams HW: 14.1, 14.42

6. Calculating RatesA) Average Rate • The speed of a rxn for a given time interval • When plotted as concentration over time, it looks like a straight line for each time interval • With many intervals, looks like a smooth curve • Note: molarity of species A is represented by [A]

7. Ex) A → B • Avg. Rate = -Δ[A]/Δt = +Δ[B]/Δt = final M – initial M final t – initial t • Note: Both will be positive in end • Units: M/s or mol/L•s or Ms-1 • What happens to rate with time?

8. B) Instantaneous Rate • The speed of a rxn at a specific time rather than an interval • Draw a straight line at the point (called a tangent) • Take slope of tangent to get ratio of Δ[A] to Δt • This represents the instantaneous rate • Concentration vs. Time Plots

9. C) Rate Stoichiometry • If not a 1:1 ratio of reactants to products, we must take into account the coefficients • Ex) 2HI(g) → H2(g) + I2(g) • Rate = -1 Δ[HI] = Δ[H2] = Δ[I2] 2 Δt Δt Δt • In general: • aA + bB → cC + dD • Rate = -1 Δ[A] = -1 Δ[B] = 1 Δ[C] = 1 Δ[D] a Δt b Δt c Δt d Δt

10. Ex) 2N2O5→ 4NO2 + O2 • If the rate of decomposition of dinitrogen pentoxide is 4.27 x 10-7 Ms-1, what is the rate of appearance of NO2? HW: 14.4, 14.10, 14.12

11. III) Rate LawsA) Definitions • Rate Law- mathematical expression that shows how rate depends on concentration; takes into account exponentially changing rates • Rate Constant- k; relates concentration to rate • Reaction Order- exponents of concentrations of reactants; determined using experimental data NOT from stoichiometry of equation; usually 0, 1, 2 but can be fraction or negative • Overall Order- sum of all exponents

12. B) Orders • Ex 1) A + B → C • Rate = k [A]m[B]n • A is • B is • Overall order is • Ex 2)NH4++ NO2-→ N2 + 2H2O • Rate = k [NH4+][NO2-] • NH4+ is • NO2- is • Overall order is • Units of k are

13. Ex 3) CHCl3 + Cl2→ CCl4 + HCl • Rate = k [CHCl3] [Cl2]1/2 • CHCl3 is • Cl2 is • Overall order is • Units of k are HW: 14.14

14. C) Using Initial Concentrations to Experimentally Determine Rate Laws

16. Experiment Number [A](M) [B](M) Initial Rate (M/s) 1 0.100 0.100 4.0  x 10–5 2 0.100 0.200 8.0  x 10–5 3 0.200 0.100 16.0 x 10–5 Example • Find where only one reactant is changing (controlled experiment) and see how it affects the rate

17. Now find the rate constant value • Pick one of the trials, and plug-n- chug! (or do all and take avg.) HW: 14.16, 14.18, 14.20, 14.22, 14.24

18. IV. The Changing of Concentration with Time • We want to have an equation that allows us to determine the concentration at any time • We won’t deal with calculus in this class, so we won’t worry about how to derive these equations. Still, you need to know whichequation goes with which order reaction

19. A) First-Order Rxns • The rate doubles as the reactant concentration doubles. • t= given or final time (sec) • 0= initial time (sec) In format, (y =  mx + b):

20. Example • The first-order rate constant for the decomposition of N2O5 is 6.82 x 10-3 s-1. If we start with 0.0300 moles of N2O5 in 2.5 L, how many moles of N2O5 will remain after 2.5 minutes?

21. B) Half-Life • A process that represents the time it takes for the concentration of the reactant to decrease by half • Independent of the initial amount for first-order rxns

22. Example • What is the half-life of N2O5?

23. C) Second-Order Rxns • Rate depends on the reactant concentration squared or two reactants each to the first power

24. Example • The reaction A --> products is second order in A. Initially [A]0 = 1.00 M; and after 25 mins, [A] = 0.25 M. What is the rate constant for this reaction? HW: 14.25, 14.26, 14.28, 14.30, 14.32

25. Kinetics Part 2 Images from: www.scielo.br/www.chem.ufl.edu/www.files.chem.vt.edu/ www.sciencecollege.co.uk/ www.rirecyclingclub.org/www.sparknotes.com/ www.huntsvilleminorlacrosse.com/ www.ucar.edu/rubyslippersbride.blogspot.com/ www.independent.co.uk/www.lifelounge.com/ www.webelements.com/www.ecvv.com/www.amazingrust.com/ www.white-hat-web-design.co.uk

26. Svante Arrhenius The Arrhenius EquationA) Background • According to the collision model of kinetics, the more collisions there are, the faster the rxn • In 1888, Swedish chemist Svante August Arrhenius proposed the idea that molecules need to have an effective collision in order for a rxn to proceed

27. An effective collision is one in which molecules collide with: • The proper orientation • The minimum amount of energy required to initiate the rxn (activation energy or Ea)

29. The value of Ea depends on the particular rxn • It is always positive • Can also be thought of as the energy required to produce the highest energy arrangement of particles in the mechanism (activated complex or transition state) which is the top of the “hill” • At higher temps, there is a greater chance that molecules will possess the Ea, thus increasing the rxn rate HW: 14.38

30. B) The Maths • f = e-Ea/RT where • f is the fraction of molecules that have an energy equal to or greater than the activation energy • e is some letter from math that represents something • Ea is the activation energy in J/mol • R is the gas constant (special guest appearance in this unit!) written as 8.314 J/mol·K • T is the temperature in K

31. Arrhenius incorporated the previous equation with the number of collisions/sec and the fraction of collisions with the correct orientation to obtain his equation: • k = Ae-Ea/RT where • k is the rate constant (notice it is dependent on temperature) • A is the frequency factor constant which takes into account collision frequency and the probability that the collisions are favorably oriented • As Ea increases, k becomes smaller

32. HW: 14.40

33. Dancing Cat Break

34. Reaction MechanismsA) Background • Series of steps taken to get to the final products • Relevant to kinetics because each step ultimately affects the overall rate • Represented by a series of equations that should “cancel out” to the final equation • The net equation is like the “before” and “after” snapshots, but tells us nothing of the actual mechanism which must be determined experimentally

35. B) Elementary Steps • Any single step process • # of molecules involved in an elementary step is called the molecularity • Unimolecular = • AB  A + B • Bimolecular = • A + B  AB • Termolecular = (very rare and unlikely, but possible) • A + B + C  ABC

36. C) Multistep Mechanisms • Involve intermediates = temporary products that are consumed in subsequent steps • 1) NO2 + NO2 NO3 + NO • 2) NO3 + CO  NO2 + CO2 • What is the intermediate? • What is the chemical equation?

37. You Try: • 1) O3 O2 + O • 2) O3 + O  2O2 • Describe the molecularity of each step. • Write the overall equation. • Identify any intermediates. HW: 14.54

38. How mechanisms affect rate lawsA) Elementary Steps • The rate laws of the elementary steps determine the overall rate law • The molecularity of each elementary step determines its order • Unimolecular is 1st order • Bimolecular is 2nd order • Termolecular is 3rd order • Thus, although the stoichiometry of the overall rxn does not determine the rate law, the stoichiometry of the elementary steps does!

39. HW: 14.56

40. B) Mechanisms with a Slow Initial Step • The slowest step in a rxn is called the rate-determining step because it determines the final rate law • Given: NO2 + CO  NO + CO2 • If this occurred as a single bimolecular elementary step you would expect the rate law to be: • Rate = k

41. But the actual observed rate law is: • Rate = k [NO2]2 • We can now propose a mechanism that would lead to this experimental rate law: • 1) NO2 + NO2 NO3 + NO (slow) • 2) NO3 + CO  NO2 + CO2 (fast) • Since Step 1 is slow, it limits how fast Step 2 can go. Thus, Step 1 is the rate-determining step and is all we need to get the rate law…

42. C) Mechanisms with a Fast Initial Step • Given: • 2NO + Br2 2NOBr • Experimental Rate Law: • Rate = k [NO]2[Br2] • Proposed Mechanism: • NO + NO + Br2 2NOBr • Why is this an unlikely mechanism?

43. Alternative mechanism: • 1) NO + Br2⇄ NOBr2 (fast) • 2) NOBr2 + NO  2NOBr (slow) • Step 2 is now rate-determining: • Rate = k2 [NOBr2][NO] • Why is this rate law unacceptable?

44. Since Step 1 is an equilibrium, the rate of the forward rxn = rate of the reverse rxn • NO + Br2 NOBr2 • Rate = k1 [NO][Br2] • NOBr2  NO + Br2 • Rate = k-1 [NOBr2] • k1 [NO][Br2] = k-1 [NOBr2] • Solve for [NOBr2] so we can substitute for it in our rate law

46. The moral of the story: when a fast step precedes a slow step, solve for the concentration of the intermediate by assuming that an equilibrium is established in the fast step! HW: 14.60, 14.82

47. Catalysts • A substance that increases the speed of a rxn without being consumed in the rxn; lowers Ea by allowing alternate rxn pathway

48. A) Homogeneous Catalysis • The catalyst is in the same phase as the reactants • Ex) KClO3(s)  KCl(s) + O2(g)

49. B) Heterogeneous Catalysis • The catalyst is in a different phase than the reactants • Active Site = place where rxn will occur • Adsorption = binding of molecules to a surface • Reactants must adsorb to an active site on the catalyst for the rxn to proceed • Video

50. C) Enzymes • Biological catalysts • A substrate binds to the enzyme in the lock and key model • The enzyme-substrate complex creates products then releases them • Inhibitors bind to a substrate site of enzyme to prevent binding (nerve poisons and toxic metal ions) HW: 14.64