1. Kinetics Chapter 15 E-mail: benzene4president@gmail.combenzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

2. Kinetics – ch 15 1. The average rate of disappearance of ozone in the reaction 2 O 3 (g) → 3 O 2 (g) is found to be 8.9 × 10 –3 atm/min. What is the rate of appearance of O 2 during this interval?

3. Kinetics – ch 15 2. The rate law for the following reaction 2NO(g) + O 2 (g)  2NO 2 (g) was experimentally found to be in the form: rate=k[NO] x [O 2 ] y It was also found that when the NO concentration was doubled, the rate of the reaction increases by a factor of 4. In addition, when both the O 2 and the NO concentration were doubled, the rate increases by a factor of 8. What is the reaction order of O 2 ?

4. Kinetics – ch 15 3. Consider the following reaction: A + 3 B  2 C a. The following data was collected for this reaction at 25  C. What is the rate law? b. What is the overall order for the reaction? c. Calculate the rate constant? d. What is the rate for experiment 4? Experiment[A] 0 [B] 0 Initial Rate (M/s) 10.00190.0310.141 20.00190.00310.0141 30.0190.03114.10 40.650.24?

5. Kinetics – ch 15 4. Given the following data for the reaction: 2 A + B  2 C Determine the rate law, the rate constant and the missing concentrations. ExperimentInitial [A]Initial [B]Initial Rate (M/min) 10.150.720.00084 20.150.360.00021 30.600.360.00021 40.85?0.0046

6. Kinetics – ch 15 5. Consider the following reaction: Cl 2 (g) + CHCl 3 (g)  HCl (g) + CCl 4 (g) The rate law for this reaction has been observed to be: Rate = k[Cl 2 ] 1/2 [CHCl 3 ] What are the units of the rate constant assuming time in seconds?

7. Kinetics – ch 15 6. What would you have to plot in order to get a straight line for the following: a. Zero order b. First order c. Second order

8. Kinetics – ch 15 7. Consider the decomposition of nitrogen dioxide: 2 NO 2  2 NO + O 2 Rate = k[NO 2 ] 2 a. What is the integrated rate law? b. A plot of 1/[NO 2 ] verses time yielded a straight line with a slope of 1.8 x 10 -3 Lmol -1 s -1 at 500 K. If the initial concentration is 1.2 M, how long will it take for the [NO 2 ] to decrease by 38%?

9. Kinetics – ch 15 8. The thermal decomposition of phosphine (PH 3 ) into phosphorus and hydrogen is a first order reaction. The half-life for this reaction is 35 sec at 680  C. a. Calculate the time required for 95% of the phosphine to decompose. b. What percentage of phosphine reacted after 78 seconds? c. What is the concentration of the hydrogen after 78 s if you started with 0.86M PH 3 ?

10. Kinetics – ch 15 9. What happens to the half-lives for zero, first and second order? a. Get longer b. Get shorter c. Stays constant What is the 3 rd half-life for a second order reaction with k = 0.22 hr -1 M -1 with an initial concentration of 12 M.

11. Kinetics – ch 15 10. For the reaction A  products, successive half-lives are observed to be 20.0, 10.0, and 5.0 min for an experiment in which [A] 0 = 0.1 M. What is the concentration of A at 38.0 min?

12. Kinetics – ch 15 11. Consider two reaction vessels, one containing A and the other containing B, with equal initial concentrations. If both substances decompose by first-order kinetics where k a = 4.5 x 10 -4 s -1 and k b = 3.7 x 10 -3 s -1 ; how much time must pass to reach a condition such that [A] = 4.00[B]?

13. Kinetics – ch 15 12. The rate of the reaction O + NO 2  NO + O 2 was studied at a certain temperature. a. In one experiment, [NO 2 ] was in large excess at a concentration of 1.0 x 10 13 molecules/cm 3 when the following data collected. What is the order with respect to the oxygen atoms? b. The reaction is known to be first order with respect to NO 2. Determine the overall rate law and the rate constant. Time (s) [O] (atoms/cm 3 ) 05.0 x 10 9 1.0 x 10 -2 1.9 x 10 9 2.0 x 10 -2 6.8 x 10 8 3.0 x 10 -2 2.5 x 10 8

14. Kinetics – ch 15 13. Given the following mechanism: H 2 O 2  H 2 O + O O + CF 2 Cl 2  ClO + CF 2 Cl ClO + O 3  Cl + 2 O 2 Cl + CF 2 Cl  CF 2 Cl 2 a. Write the overall equation for the reaction? b. Identify the reaction intermediates. c. Identify the catalyst.

15. Kinetics – ch 15 14. Write the overall reaction, identify any intermediates and derive a rate law given the following reaction mechanism: Cl 2 ⇌ 2 Cl (fast equilibrium) Cl + CHCl 3  HCl + CCl 3 (slow) Cl + CCl 3  CCl 4 (fast)

16. Kinetics – ch 15 15. The following mechanism is proposed for the reduction of NO 3 - by MoCl 6 2- : MoCl 6 2- k1k1 k -1 MoCl 5 - + Cl - NO 3 - + MoCl 5 - k2k2 OMoCl 5 - + NO 2 - a. What is the intermediate? b. Derive an expression for the rate law (rate = d[NO 2 - ]/dt) for the overall reaction using steady-state approximation.

17. Kinetics – ch 15 16. The forward activation energy for an elementary step is 42 kJ/mol and the reverse activation energy is 32 kJ/mol. Calculate  E  for the step.

18. Kinetics – ch 15

19. 18. The activation energy for the reaction H 2 (g) + I 2 (g) → 2 HI (g) is changed from 184 kJ/mol to 59.0 kJ/mol both at 600. K by the introduction of a Pt catalyst. Calculate the ratio of the catalyzed rate constant to the un-catalyzed rate constant. Assume A is constant.

20. Kinetics – ch 15 You have completed ch. 15

21. Ch 15 – Answer Key

22. 2. The rate law for the following reaction 2NO(g) + O 2 (g)  2NO 2 (g) was experimentally found to be in the form: rate=k[NO] x [O 2 ] y It was also found that when the NO concentration was doubled, the rate of the reaction increases by a factor of 4. In addition, when both the O 2 and the NO concentration were doubled, the rate increases by a factor of 8. What is the reaction order of O 2 ? k is always constant and if O 2 is also constant the rate law becomes ⇒ rate α [NO] x ⇒ 4 α 2 x ⇒ x = 2 now if O 2 is not constant the rate law is rate α [NO] 2 [O 2 ] y ⇒ 8 α [2] 2 [2] y ⇒ y = 1

23. Ch 15 – Answer Key

24. 3. continued b. 3 c. since rate=[A] 2 [B] 1 ⇒ using data from expt 1 ⇒ 0.141M/s=k(0.0019M) 2 (0.031M) ⇒ k = 1.26x10 6 M -2 s -1 d. since rate=[A] 2 [B] 1 ⇒ rate=(1.26x10 6 M -2 s -1 )(0.65M) 2 (0.24M)= 1.28x10 5 M/s

25. Ch 15 – Answer Key

26. 5. Consider the following reaction: Cl 2 (g) + CHCl 3 (g)  HCl (g) + CCl 4 (g) The rate law for this reaction has been observed to be: Rate = k[Cl 2 ] 1/2 [CHCl 3 ] What are the units of the rate constant assuming time in seconds? units for k = t -1 M 1-overall order The overall order for this reaction is 1.5 ⇒ the units for k ⇒ s -1 M -0.5

27. Ch 15 – Answer Key 6. What would you have to plot in order to get a straight line for the following: a. Zero order ⇒ conc. vs time ⇒ k = -slope b. First order ⇒ ln(conc.) vs time ⇒ k = -slope c. Second order ⇒ 1/conc. vs time ⇒ k = slope

28. Ch 15 – Answer Key

29. 8. The thermal decomposition of phosphine (PH 3 ) into phosphorus and hydrogen is a first order reaction. The half-life for this reaction is 35 sec at 680  C. a. Calculate the time required for 95% of the phosphine to decompose. for 1 st order reactions ⇒ t 1/2 =0.693/k ⇒ k = 0.693/35s ⇒ k = 0.0198s-1 if 95% has decomposed there’s 5% left ln(5) = -(0.0198)(t) + ln(100) ⇒ t = 151s NOTE ⇒ %s can only be used in 1 st order rate laws b. What percentage of phosphine reacted after 78 seconds? ln[A] = -(0.0198)(78s) + ln(100) ⇒ [A] = 21.3 ⇒ 21.3% remains so 78.7% reacted

30. Ch 15 – Answer Key 8. continued c. What is the concentration of the hydrogen after 78 s if you started with 0.86M PH 3 ? the balanced reaction is 2 PH 3  2 P + 3 H 2 2 PH 3  2 P3 H 2 0.8600 -(0.86)(0.787)+(0.86)(0.787)+(3/2)(0.86)(0.787) 0.180.6771.02 from part b 78.7% reacts after 78 s molar ratio from coefficients

31. Ch 15 – Answer Key

33. 11. Consider two reaction vessels, one containing A and the other containing B, with equal initial concentrations. If both substances decompose by first- order kinetics where k a = 4.5 x 10 -4 s -1 and k b = 3.7 x 10 -3 s -1 ; how much time must pass to reach a condition such that [A] = 4.00[B]? since they are both 1 st order ⇒ ln[A]=-k a t + ln[A] 0 and ln[B] = -k b t + ln[B] 0 ⇒ since they have the same initial concentrations ⇒ ln[A] + k a t = ln[B] + k b t since [A]=4[B] ⇒ ln4[B] + k a t = ln[B] + k b t ⇒ ln4 = k b t – k a t ⇒ ln4 = t(3.7 x 10 -3 s -1 - 4.5 x 10 -4 s -1 ) ⇒ t = 427s

34. Ch 15 – Answer Key

35. b. since both O and NO 2 are 1 st order the rate law is ⇒ rate = k[O][NO 2 ] ⇒ however when the concentration of one substance is really high you can write a pseudo rate law ⇒ rate = k’[O] ⇒ where k’ is the -slope of the line ⇒ setting the two rate laws equal to each other ⇒ k[O][NO 2 ] = k’[O] or k[NO 2 ] = k’ ⇒ k(1.0 x 10 13 molecules/cm 3 ) = 99s-1 ⇒ k = 9.9x10 -12 s -1 molecules -1 cm 3

36. Ch 15 – Answer Key 13. Given the following mechanism: H 2 O 2  H 2 O + O O + CF 2 Cl 2  ClO + CF 2 Cl ClO + O 3  Cl + 2 O 2 Cl + CF 2 Cl  CF 2 Cl 2 a. Write the overall equation for the reaction? sum of the elementary steps ⇒ if you add them up the O, CF 2 Cl 2, ClO, CF 2 Cl and Cl cancel out leaving you with H 2 O 2 + O 3  H 2 O + 2 O 2 b. Identify the reaction intermediates. ⇒ O, ClO, CF 2 Cl and Cl ⇒ reaction intermediates are temporarily made then consumed c. Identify the catalyst. ⇒ CF 2 Cl 2 ⇒ catalysts speed up the reaction by changing the pathway and lowering the activation energy ⇒ catalysts are temporarily consumed then regenerated

37. Ch 15 – Answer Key

39. 16. The forward activation energy for an elementary step is 42 kJ/mol and the reverse activation energy is 32 kJ/mol. Calculate  E  for the step. ΔE°= forward E a – reverse E a ΔE°= 42kJ/mol – 32kJ/mol = 10kJ/mol

40. Ch 15 – Answer Key