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. I.What you (maybe) already know about kinetics A) Definitions What is a reaction rate? What is a reaction rate? What is a reaction mechanism? What is a reaction mechanism?

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

4. C) Factors Affecting Rxn Rate Factor Effect On Rate Why? 1. Catalyst 2. Inc. Temp. 3. Inc. Surface Area 4. Inc. Concentration 5. Inc. Pressure

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

6. II.Calculating Rates A) Average Rate The speed of a rxn for a given time interval 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 When plotted as concentration over time, it looks like a straight line for each time interval With many intervals, looks like a smooth curve With many intervals, looks like a smooth curve Note: molarity of species A is represented by [A] 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 Avg. Rate = -Δ[A]/Δt = +Δ[B]/Δt = final M – initial M final t – initial t Note: Both will be positive in end Note: Both will be positive in end Units: M/s or mol/Ls or Ms -1 Units: M/s or mol/Ls or Ms -1 What happens to rate with time? What happens to rate with time?

8. B) Instantaneous Rate The speed of a rxn at a specific time rather than an interval The speed of a rxn at a specific time rather than an interval Draw a straight line at the point (called a tangent) Draw a straight line at the point (called a tangent) Take slope of tangent to get ratio of Δ[A] to Δt Take slope of tangent to get ratio of Δ[A] to Δt This represents the instantaneous rate This represents the instantaneous rate Concentration vs. Time Plots Concentration vs. Time Plots Concentration vs. Time Plots 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 If not a 1:1 ratio of reactants to products, we must take into account the coefficients Ex) 2HI(g) → H 2 (g) + I 2 (g) Ex) 2HI(g) → H 2 (g) + I 2 (g) Rate = -1 Δ[HI] = Δ[H 2 ] = Δ[I 2 ] 2 Δt Δt Δt Rate = -1 Δ[HI] = Δ[H 2 ] = Δ[I 2 ] 2 Δt Δt Δt In general: In general: aA + bB → cC + dD 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 Rate = -1 Δ[A] = -1 Δ[B] = 1 Δ[C] = 1 Δ[D] a Δt b Δt c Δt d Δt

10. Ex) 2N 2 O 5 → 4NO 2 + O 2 If the rate of decomposition of dinitrogen pentoxide is 4.27 x 10 -7 Ms -1, what is the rate of appearance of NO 2 ? If the rate of decomposition of dinitrogen pentoxide is 4.27 x 10 -7 Ms -1, what is the rate of appearance of NO 2 ? HW: 14.4, 14.10, 14.12

11. III) Rate Laws A) Definitions Rate Law- mathematical expression that shows how rate depends on concentration; takes into account exponentially changing rates Rate Law- mathematical expression that shows how rate depends on concentration; takes into account exponentially changing rates Rate Constant- k; relates concentration to rate 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 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 Overall Order- sum of all exponents

12. B) Orders Ex 1) A + B → C Ex 1) A + B → C Rate = k [A] m [B] n Rate = k [A] m [B] n A is A is B is B is Overall order is Overall order is Ex 2)NH 4 + + NO 2 - → N 2 + 2H 2 O Ex 2)NH 4 + + NO 2 - → N 2 + 2H 2 O Rate = k [NH 4 + ][NO 2 - ] Rate = k [NH 4 + ][NO 2 - ] NH 4 + is NH 4 + is NO 2 - is NO 2 - is Overall order is Overall order is Units of k are Units of k are

13. Ex 3) CHCl 3 + Cl 2 → CCl 4 + HCl Ex 3) CHCl 3 + Cl 2 → CCl 4 + HCl Rate = k [CHCl 3 ] [Cl 2 ] 1/2 Rate = k [CHCl 3 ] [Cl 2 ] 1/2 CHCl 3 is CHCl 3 is Cl 2 is Cl 2 is Overall order is Overall order is Units of k are Units of k are HW: 14.14

14. C) Using Initial Concentrations to Experimentally Determine Rate Laws If the order of a particular reactant is… …the effect on the rate if its concentration is doubled is… 0 1 2 3

15. If the order of a particular reactant is… …the effect on the rate if its concentration is tripled is… 0 1 2 3

16. Example Find where only one reactant is changing (controlled experiment) and see how it affects the rate Find where only one reactant is changing (controlled experiment) and see how it affects the rate Experiment Number[A](M)[B](M) Initial Rate (M/s) 10.100 4.0 x 10 –5 20.1000.2008.0 x 10 –5 30.2000.10016.0 x 10 –5

17. Now find the rate constant value Now find the rate constant value Pick one of the trials, and plug-n- chug! (or do all and take avg.) 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 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 which equation goes with which order reaction 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 which equation goes with which order reaction

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

20. Example The first-order rate constant for the decomposition of N 2 O 5 is 6.82 x 10 -3 s -1. If we start with 0.0300 moles of N 2 O 5 in 2.5 L, how many moles of N 2 O 5 will remain after 2.5 minutes? The first-order rate constant for the decomposition of N 2 O 5 is 6.82 x 10 -3 s -1. If we start with 0.0300 moles of N 2 O 5 in 2.5 L, how many moles of N 2 O 5 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 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 Independent of the initial amount for first-order rxns

22. Example What is the half-life of N 2 O 5 ? What is the half-life of N 2 O 5 ?

23. C) Second-Order Rxns Rate depends on the reactant concentration squared or two reactants each to the first power 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? 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. V.The Arrhenius Equation A) Background According to the collision model of kinetics, the more collisions there are, the faster the rxn 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 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 Svante Arrhenius

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

29. The value of E a depends on the particular rxn The value of E a depends on the particular rxn It is always positive 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” 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 E a, thus increasing the rxn rate At higher temps, there is a greater chance that molecules will possess the E a, thus increasing the rxn rate HW: 14.38

30. B) The Maths f = e -Ea/RT where f = e -Ea/RT where f is the fraction of molecules that have an energy equal to or greater than the activation energy 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 e is some letter from math that represents something E a is the activation energy in J/mol E a 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 R is the gas constant (special guest appearance in this unit!) written as 8.314 J/mol·K T is the temperature in 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: 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 = Ae -Ea/RT where k is the rate constant (notice it is dependent on temperature) 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 A is the frequency factor constant which takes into account collision frequency and the probability that the collisions are favorably oriented As E a increases, k becomes smaller As E a increases, k becomes smaller

32. HW: 14.40

33. Dancing Cat Break

34. VI.Reaction Mechanisms A) Background Series of steps taken to get to the final products Series of steps taken to get to the final products Relevant to kinetics because each step ultimately affects the overall rate 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 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 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 Any single step process # of molecules involved in an elementary step is called the molecularity # of molecules involved in an elementary step is called the molecularity Unimolecular = Unimolecular = AB  A + B AB  A + B Bimolecular = Bimolecular = A + B  AB A + B  AB Termolecular = (very rare and unlikely, but possible) Termolecular = (very rare and unlikely, but possible) A + B + C  ABC A + B + C  ABC

36. C) Multistep Mechanisms Involve intermediates = temporary products that are consumed in subsequent steps Involve intermediates = temporary products that are consumed in subsequent steps 1) NO 2 + NO 2  NO 3 + NO 1) NO 2 + NO 2  NO 3 + NO 2) NO 3 + CO  NO 2 + CO 2 2) NO 3 + CO  NO 2 + CO 2 What is the intermediate? What is the intermediate? What is the chemical equation? What is the chemical equation?

37. You Try: You Try: 1) O 3  O 2 + O 1) O 3  O 2 + O 2) O 3 + O  2O 2 2) O 3 + O  2O 2 Describe the molecularity of each step. Describe the molecularity of each step. Write the overall equation. Write the overall equation. Identify any intermediates. Identify any intermediates. HW: 14.54

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

39. Molecularity Elementary Step Rate Law Unimolecular A  products Bimolecular A + A  products Bimolecular A + B  products Termolecular A + A + A  products Termolecular A + A + B  products Termolecular A + B + C  products 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 The slowest step in a rxn is called the rate-determining step because it determines the final rate law Given: NO 2 + CO  NO + CO 2 Given: NO 2 + CO  NO + CO 2 If this occurred as a single bimolecular elementary step you would expect the rate law to be: If this occurred as a single bimolecular elementary step you would expect the rate law to be: Rate = k Rate = k

41. But the actual observed rate law is: But the actual observed rate law is: Rate = k [NO 2 ] 2 Rate = k [NO 2 ] 2 We can now propose a mechanism that would lead to this experimental rate law: We can now propose a mechanism that would lead to this experimental rate law: 1) NO 2 + NO 2  NO 3 + NO (slow) 1) NO 2 + NO 2  NO 3 + NO (slow) 2) NO 3 + CO  NO 2 + CO 2 (fast) 2) NO 3 + CO  NO 2 + CO 2 (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… 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: Given: 2NO + Br 2  2NOBr 2NO + Br 2  2NOBr Experimental Rate Law: Experimental Rate Law: Rate = k [NO] 2 [Br 2 ] Rate = k [NO] 2 [Br 2 ] Proposed Mechanism: Proposed Mechanism: NO + NO + Br 2  2NOBr NO + NO + Br 2  2NOBr Why is this an unlikely mechanism? Why is this an unlikely mechanism?

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

44. Since Step 1 is an equilibrium, the rate of the forward rxn = rate of the reverse rxn Since Step 1 is an equilibrium, the rate of the forward rxn = rate of the reverse rxn NO + Br 2  NOBr 2 NO + Br 2  NOBr 2 Rate = k 1 [NO][Br 2 ] Rate = k 1 [NO][Br 2 ] NOBr 2  NO + Br 2 NOBr 2  NO + Br 2 Rate = k -1 [NOBr 2 ] Rate = k -1 [NOBr 2 ] k 1 [NO][Br 2 ] = k -1 [NOBr 2 ] k 1 [NO][Br 2 ] = k -1 [NOBr 2 ] Solve for [NOBr 2 ] so we can substitute for it in our rate law Solve for [NOBr 2 ] 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! 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. VIII.Catalysts A substance that increases the speed of a rxn without being consumed in the rxn; lowers E a by allowing alternate rxn pathway A substance that increases the speed of a rxn without being consumed in the rxn; lowers E a by allowing alternate rxn pathway

48. A) Homogeneous Catalysis The catalyst is in the same phase as the reactants The catalyst is in the same phase as the reactants Ex) KClO 3 (s)  KCl(s) + O 2 (g) Ex) KClO 3 (s)  KCl(s) + O 2 (g)

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

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

51. Stop! Stop!

52. Dancing Cat Finale