1. 1 Kinetics Chapter 15

2. 2 The study of rxn rates Rxn rate =  concentration/  time Rxn rate =  concentration/  time Example: Example: 2N 2 O 5  4NO 2 + O 2 As reactant concentration decreases  products’ concentrations increase As reactant concentration decreases  products’ concentrations increase

3. 3 Let’s do some calculations From 300s to 400s, From 300s to 400s, -  [N 2 O 5 ]/  t = -.0019M/100s What are  [NO 2 ]/  t &  [O 2 ]/  t during the same time interval? What are  [NO 2 ]/  t &  [O 2 ]/  t during the same time interval? Let’s work this out… Let’s work this out… Is rxn rate constant? Is rxn rate constant? –Let’s look at  [NO 2 ]/  t at different time intervals

4. 4 More… The previous rates were average rates The previous rates were average rates Different from instantaneous rates (@ single pt) Different from instantaneous rates (@ single pt) Ex:  [N 2 O 5 ]=-0.0019M ( @ 200s) Ex:  [N 2 O 5 ]=-0.0019M ( @ 200s) Car speed is analogous Car speed is analogous

5. 5 Continued We can also calculate changes mathematically based on stoichiometric relationships We can also calculate changes mathematically based on stoichiometric relationships Let’s see Let’s see http://wps.prenhall.co m/wps/media/objects /167/172009/Decomp ositionofN2O5.html http://wps.prenhall.co m/wps/media/objects /167/172009/Decomp ositionofN2O5.html http://wps.prenhall.co m/wps/media/objects /167/172009/Decomp ositionofN2O5.html http://wps.prenhall.co m/wps/media/objects /167/172009/Decomp ositionofN2O5.html

6. 6 Writing rate expression 2N 2 O 5  4NO 2 + O 2 To equate rates of disappearance or appearance To equate rates of disappearance or appearance –Divide the stoichiometric coefficient in the balanced equation

7. 7 Let’s work on this a) Give four related rate expressions for the rate of the following reaction: a) Give four related rate expressions for the rate of the following reaction: 2H 2 CO (g) + O 2(g)  2CO (g) + 2H 2 O (g) b) Give three related rate expressions for the rate of the following reaction: b) Give three related rate expressions for the rate of the following reaction: N 2(g) + 3H 2(g)  2NH 3(g)

8. 8 Let’s also work on this: Consider the reaction: Consider the reaction: A + 2B  C A + 2B  C Give the three related rate expressions for the rate of the reaction. Give the three related rate expressions for the rate of the reaction. Using the following data, determine the average rate of the reaction, and the rate between 30 and 40 seconds. Using the following data, determine the average rate of the reaction, and the rate between 30 and 40 seconds. Time (s): 0.0, 10.0, 20.0, 30.0, & 40.0 Time (s): 0.0, 10.0, 20.0, 30.0, & 40.0 [A] (M): 1.000, 0.833, 0.714, 0.625, & 0.555, respectively [A] (M): 1.000, 0.833, 0.714, 0.625, & 0.555, respectively

9. 9 Reaction conditions and rates Higher temp  faster rxn rate Higher temp  faster rxn rate –Raise in temp by 10°C  double reaction rate Higher concentration  faster rxn rate Higher concentration  faster rxn rate Catalysts speed up rxn rates Catalysts speed up rxn rates –They don’t react –They lower the activation energy Enzymes (proteins) in organisms Enzymes (proteins) in organisms Metals, salts, etc. in chemical rxns Metals, salts, etc. in chemical rxns Question: What does acid rain do to enzymes? Question: What does acid rain do to enzymes?

10. 10 Concentration: 2N 2 O 5  4NO 2 + O 2 If you double [N 2 O 5 ], you double the rxn rate If you double [N 2 O 5 ], you double the rxn rate –Rate of rxn  [N 2 O 5 ] Other rxns have different relationships w/concentration-rxn rates Other rxns have different relationships w/concentration-rxn rates

11. 11 Rate equations Describe relationship between reactant concentration and rxn rate Describe relationship between reactant concentration and rxn rate For 2N 2 O 5  4NO 2 + O 2 For 2N 2 O 5  4NO 2 + O 2 –Rate of rxn = k[N 2 O 5 ] k = rate constant k = rate constant –Is the rate of rxn constant too? Generic expression for rate eq: aA + bB  xX Generic expression for rate eq: aA + bB  xX So rate = k[A] m [B] n So rate = k[A] m [B] n –Where m & n need not equal stoichiometric ratios! –They can be zero, fractions, even negative #’s –Empirically verified values

12. 12 Order of rxn: Exponent of its concentration term for each item 2NO (g) + Cl 2(g)  2NOCl (g) 2NO (g) + Cl 2(g)  2NOCl (g) Rxn rate for NO is second order Rxn rate for NO is second order Rxn rate for Cl 2 is first order Rxn rate for Cl 2 is first order –Thus, rxn rate = k[NO] 2 [Cl 2 ] Total order of rxn = summation of exponents of all concentration items Total order of rxn = summation of exponents of all concentration items –Therefore, overall rxn rate is third order Remember, m & n need not equal stoichiometric ratios! Remember, m & n need not equal stoichiometric ratios!

13. 13 Reaction mechanisms Step by step chemical equations that, when summed, give net rxn Step by step chemical equations that, when summed, give net rxn For ex: For ex: Br 2(g) + 2NO (g)  2BrNO (g) 1) Br 2(g) + NO (g)  Br 2 NO (g) 2) Br 2 NO (g) + NO (g)  2BrNO (g) Each step: Each step: –Elementary step

14. 14 Molecularity Elementary steps classified according to # of reactant molecules Elementary steps classified according to # of reactant molecules –Molecularity Unimolecular: Unimolecular: –A  products –Rate expression: k[A] Bimolecular: Bimolecular: –A + B  products –Rate expression: k[A][B] Termolecular: Termolecular: –A + B + C  products (Or 3A or 2A + B, etc.) –Rate expression: k[A][B][C] (or k[A] 3 or k[A] 2 [B], etc.) In elementary steps, stoichiometry defines rate equation! In elementary steps, stoichiometry defines rate equation! –K different for each step

15. 15 Rate-determining step The slowest elementary step is the rate- limiting step The slowest elementary step is the rate- limiting step –Overall rxn follows this rate-limiting step However, intermediate not included in final rate law However, intermediate not included in final rate law

16. 16 An example 2NO (g) + O 2(g) → 2NO 2(g) NO (g) + NO (g)  N 2 O 2(g) (fast, eq.) NO (g) + NO (g)  N 2 O 2(g) (fast, eq.) N 2 O 2(g) + O 2(g) → NO 2(g) (slow) N 2 O 2(g) + O 2(g) → NO 2(g) (slow) Using slow, rate-determining, step Using slow, rate-determining, step –Rate = k 2 [N 2 O 2 ][O 2 ] Since fast-step at eq. Since fast-step at eq. –k 1 [NO] 2 = k -1 [N 2 O 2 ] –Thus, K = k 1 /k -1 = [N 2 O 2 ]/[NO] 2 Intermediate ([N 2 O 2 ]) = K[NO] 2 Intermediate ([N 2 O 2 ]) = K[NO] 2 –Since rate = k 2 [N 2 O 2 ][O 2 ] Rate = k 2 K[NO] 2 [O 2 ] Rate = k 2 K[NO] 2 [O 2 ] Let’s say k 2 K = k’ Let’s say k 2 K = k’  rate = k’[NO] 2 [O 2 ]  rate = k’[NO] 2 [O 2 ]

17. 17 Another problem 2H 2(g) + 2NO (g)  2H 2 O (g) + N 2(g) 2NO (g)  N 2 O 2(g) ; fast 2NO (g)  N 2 O 2(g) ; fast H 2(g) + N 2 O 2(g)  H 2 O (g) + N 2 O (g) ; slow H 2(g) + N 2 O 2(g)  H 2 O (g) + N 2 O (g) ; slow N 2 O (g) + H 2(g)  N 2(g) + H 2 O (g) ; fast N 2 O (g) + H 2(g)  N 2(g) + H 2 O (g) ; fast

18. 18 Solution

19. 19 Rate = change 2NH 3  N 2 + 3H 2 2NH 3  N 2 + 3H 2 Rate = k[NH 3 ] 0 = k Rate = k[NH 3 ] 0 = k –Zero order rxn rate Any tinkering of concentration will not change rxn rate of species Any tinkering of concentration will not change rxn rate of species k = mol/(L  time) k = mol/(L  time)

20. 20 Rate = k[NO] 2 [Cl 2 ] Rxn rate for Cl 2 is first order Rxn rate for Cl 2 is first order  rate doubled when [Cl 2 ] doubled  rate doubled when [Cl 2 ] doubled k = time -1 k = time -1

21. 21 Rate = k[NO] 2 [Cl 2 ] Rxn rate for NO is second order Rxn rate for NO is second order Doubling [NO]  quadruples (x4) rxn rate Doubling [NO]  quadruples (x4) rxn rate k = L/(mol  time) k = L/(mol  time)

22. 22 Let’s do this: Nitrosyl bromide, NOBr, is formed from NO and Br 2. Nitrosyl bromide, NOBr, is formed from NO and Br 2. 2NO (g) + Br 2(g)  2NOBr (g) Experiment show that the reaction is first order in Br 2 and second order in NO. Experiment show that the reaction is first order in Br 2 and second order in NO. –Write the rate law for the reaction. If the concentration of Br 2 is tripled, how will the reaction rate change? If the concentration of Br 2 is tripled, how will the reaction rate change? What happens to the reaction rate when the concentration of NO is doubled? What happens to the reaction rate when the concentration of NO is doubled?

23. 23 Determining rate equations Done in the lab Done in the lab Generally, calculated after 1-2% of limiting reactant consumed Generally, calculated after 1-2% of limiting reactant consumed –Lessens chance of side-rxns throwing things off

24. 24 Let’s work on this: The rate for the oxidation of iron (II) by cerium (IV) is measured at several different initial concentrations of the two reactants: The rate for the oxidation of iron (II) by cerium (IV) is measured at several different initial concentrations of the two reactants: Ce 4+ (aq) + Fe 2+ (aq)  Ce 3+ (aq) + Fe 3+ (aq) [Ce 4+ ] (M): 1.1 x 10 -5, 1.1 x 10 -5, 3.4 x 10 -5 [Ce 4+ ] (M): 1.1 x 10 -5, 1.1 x 10 -5, 3.4 x 10 -5 [Fe 2+ ] (M): 1.8 x 10 -5, 2.8 x 10 -5, 2.8 x 10 -5 [Fe 2+ ] (M): 1.8 x 10 -5, 2.8 x 10 -5, 2.8 x 10 -5 Rate (M/s): 2.0 x 10 -7, 3.1 x 10 -7, 9.5 x 10 -7 Rate (M/s): 2.0 x 10 -7, 3.1 x 10 -7, 9.5 x 10 -7 –Write the rate law for this reaction, determining the orders of the reaction with respect to Ce (IV) and Fe (II). What is the overall order of the reaction? –Calculate the rate constant, k, and give its units. –Predict the reaction rate for a solution in which [Ce 4+ ] is 2.6 x 10 -5 M and [Fe 2+ ] is 1.3 x 10 -5 M.

25. 25 Concentration-time relationships: Integrated rate laws Zero-order rxns Zero-order rxns -(  [R]/  t) = k[R] 0 -(  [R]/  t) = k[R] 0 Using integral calculus, integrated rate equation: Using integral calculus, integrated rate equation: k = concentration/time

26. 26 Concentration-time relationships: Integrated rate laws First-order rxns First-order rxns -  [R]/  t = k[R] -  [R]/  t = k[R] k = time -1

27. 27 Integrated rate laws Second-order rxns Second-order rxns -  [R]/  t = k[R] 2 -  [R]/  t = k[R] 2 k=1/(concentration  time)

28. 28 Using graphs to solve rxn order and rate constants Zero-order: Zero-order: [R] i – [R] f = kt [R] i – [R] f = kt Rearrange this in the form: Rearrange this in the form: y = mx + b

29. 29 Using graphs to solve rxn order and rate constants First-order: First-order: Rearrange this in the form: Rearrange this in the form: y = mx + b Solve the following: 2H 2 O 2(aq)  2H 2 O (l) + O 2(g) The reaction is first order in peroxide, and the rate constant, k, for this reaction is 1.06 x 10 -3 /min. If the initial peroxide concentration is 0.020 M, what is the concentration after 135 min?

30. 30 Solution

31. 31 Using graphs to solve rxn order and rate constants Second-order: Second-order: Rearrange this in the form: Rearrange this in the form: y = mx + b The gas phase decomposition of HI into hydrogen and iodine is second order in HI. The rate constant for the reaction is 30/M  min. How long must one wait for the concentration of HI to decrease from 0.010 M to 0.0050 M?

32. 32 Solution

33. 33 Half-life Time required for reactant concentration to decrease to ½ initial value Time required for reactant concentration to decrease to ½ initial value –Longer half-life means slower rxn –Usually used for 1 st -order rxns (like radioactive decay)

34. 34 Half-life For 1 st -order rxns, t 1/2 is independent of conc For 1 st -order rxns, t 1/2 is independent of conc

35. 35 Half-life: 2 nd order rxns Half-life increases with decreasing concentration Half-life increases with decreasing concentration Derive this! Derive this!

36. 36 Half-life: zero-order rxns Half-life decreases with decreasing concentration Half-life decreases with decreasing concentration

37. 37 Problem Sucrose decomposes to fructose and glucose in acid solution with the rate law of: Sucrose decomposes to fructose and glucose in acid solution with the rate law of: Rate = k[sucrose]; k=0.208 hr -1 @ 25°C What amount of time is required for 87.5% of the initial concentration of sucrose to decompose? What amount of time is required for 87.5% of the initial concentration of sucrose to decompose?

38. 38 Solution

39. 39 Collision Theory 1. Reacting molecules must collide with one another. 2. Reacting molecules must collide with sufficient energy to break bonds. 3. Molecules must collide in correct orientation to form new species.

40. 40 Temperature Higher temperatures lead to fast reactions Higher temperatures lead to fast reactions Molecules have varying temperatures Molecules have varying temperatures –At higher temperatures, more molecules have higher energies

41. 41 Activation energy and catalysts Activation energy: Activation energy: –Energy required to form products Methane + oxygen gas Methane + oxygen gas –Low activation energy Xenon + oxygen gas Xenon + oxygen gas –Humongous activation energy Catalysts dramatically reduce activation energy w/out being consumed Catalysts dramatically reduce activation energy w/out being consumed

42. 42 Arrhenius equation Includes tenets of collision theory; i.e., collision frequency, temp (energy), and correct orientation Includes tenets of collision theory; i.e., collision frequency, temp (energy), and correct orientation k = Ae -E a /RT k = Ae -E a /RT A = frequency factor (L/mol  s) A = frequency factor (L/mol  s) –Empirically derived relationship between reaction rate and temp e -E a /RT = fraction of molecules w/min energy needed for rxn (R = gas constant = 8.314 J/mol  K) e -E a /RT = fraction of molecules w/min energy needed for rxn (R = gas constant = 8.314 J/mol  K) Can put equation in “y=mx+b” form Can put equation in “y=mx+b” form –How?

43. 43 The Arrhenius equation

44. 44 Problem Given: Given: y = -24371x + 28.204 Solve for E a Solve for E a

45. 45 Solution

46. 46 If given two points…