Kinetics Chapter 15 E-mail: [email protected] Web-site: http://clas.sa.ucsb.edu/staff/terri/ Kinetics ch 15 1. The average rate of disappearance of ozone in the reaction 2 O3(g) 3 O2(g) is found to be 8.90 103 atm/min. What is the rate of appearance of O 2 during this interval? a. 8.90 103 atm/min b. 1.78 102 atm/min c. 1.48 102 atm/min

d. 2.67 102 atm/min Kinetics ch 15 2. Determine the rate law for the following reaction: 2 NO(g) + O2(g) 2 NO2(g) 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 O2 and the NO concentration were doubled, the rate increases by a factor of 8. What is the reaction order of O2? a. Rate = k[NO]2[O2] b. Rate = k[NO]2[O2]2 c. Rate = k[NO][O2]

d. Rate = k[NO][O2]2 e. more information is required Kinetics ch 15 3. Consider the following reaction: A + 3 B 2 C a. The following data was collected for this reaction at 25C. 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) 1 0.0019 0.031 0.141

2 0.0019 0.0031 0.0141 3 0.019

0.031 14.10 4 0.65 0.24 ? Kinetics ch 15

4. Given the following data for the reaction: 2A + B 2C Determine the rate law, the rate constant and the missing concentration. Experiment Initial [A] Initial [B] Initial Rate (M/min) 1

0.15 0.72 0.00084 2 0.15 0.36

0.00021 3 0.60 0.36 0.00021 4 0.85

? 0.0046 Kinetics ch 15 5. The following data was collected for the reaction A products. a. What is the rate law for the reaction? b. What is the rate constant? c. What is the concentration of A after 31 sec? Kinetics ch 15 6. Consider the decomposition of nitrogen dioxide:

2 NO2 2 NO + O2 Rate = k[NO2]2 A plot of 1/[NO2] verses time yielded a straight line with a slope of 1.8 x 10-3 Lmol-1s-1 at 500 K. If the initial concentration is 1.2 M, how long will it take for the [NO2] to decrease by 38%? Kinetics ch 15 7. 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 680C. 4 PH3 P4 + 6 H2 a. Calculate the time required for 95% of the phosphine to decompose.

b. What percentage of phosphine reacted after 78 seconds? Kinetics ch 15 8. Consider the following elementary (single-step) reaction: 2 NO2 (g) 2 NO (g) + O2 (g) The rate constant (k) for this reaction is 0.86 M-1 s -1 at 600K. Into a rigid, evacuated container, 2.50 atm of NO2 is added. Calculate the total pressure in the container at 600K after 20 seconds. a. 3.75 atm b. 2.50 atm c. 1.33 atm d. 3.09 atm

e. 1.76 atm Kinetics ch 15 9. Based on the following data determine the order for the reactant: 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 42.0 min? 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 ka = 4.5 x 10-4 s-1 and kb = 3.7 x 10-3 s-1; how much time must pass to reach a condition such that [A] = 4.00[B]? Kinetics ch 15 12. Consider the following reaction in which HCl adds across the double bond of ethylene: HCl + H2C=CH2 H3CCH2Cl The following mechanism, with the energy diagram shown below, has been suggested for this reaction: Step 1: HCl + H2C=CH2 H3CCH2 + + Cl Step 2: H3CCH2 + + Cl H3CCH2Cl a. Based on the potential energy diagram above, which step is rate limiting? b. What is the expected overall order of the reaction based on the proposed

mechanism? c. How many transition states occur during this reaction? Kinetics ch 15 13. Given the following mechanism: H2O2 H2O + O O + CF2Cl2 ClO + CF2Cl ClO + O3 Cl + 2 O2 Cl + CF2Cl CF2Cl2 a. Write the overall equation for the reaction? b. Identify the reaction intermediates. c. Identify the catalyst.

Kinetics ch 15 14. Write the overall reaction, identify any intermediates and derive a rate law given the following reaction mechanism: Cl2 2 Cl 2 Cl (fast equilibrium) Cl + CHCl3 HCl + CCl3 Cl + CCl3 CCl4 (fast) (slow)

Kinetics ch 15 15. Derive a rate law given the following reaction mechanism: Cl2 2 Cl 2 Cl (fast equilibrium) Cl + CO 2 Cl COCl (fast equilibrium) COCl + Cl2 COCl2 + Cl (Slow) 2Cl Cl2 (Fast) 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. Kinetics ch 15 17. The half-life for a first order reaction is 28 min at 300K. What is the half-life at 600K if the activation energy is 34 kJ/mol? Kinetics ch 15 18. The activation energy for the reaction H2 (g) + I2 (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.

Kinetics ch 15 You have completed ch. 15 Ch 15 Answer Key 1. The average rate of disappearance of ozone in the reaction 2 O3(g) 3 O2(g) is found to be 8.9 103 atm/min. What is the rate of appearance of O 2 during this interval? 8.9 103 atm/min = 1.33 x 10-2 atm/min Ch 15 Answer Key 2. The rate law for the following reaction

2NO(g) + O2(g) 2NO2(g) was experimentally found to be in the form: rate=k[NO]x[O2]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 O2? k is always constant and if O2 is also constant the rate law becomes rate [NO]NO]x 4 2x x = 2 now if O2 is not constant the rate law is rate [NO]NO]2[NO]O2]y 8 [NO]2]2[NO]2]y y = 1 rate = [NO]NO]2[NO]O2] Ch 15 Answer Key 3.

Consider the following reaction: A + 3 B 2 C a. The following data was collected for this reaction at rate law? b. What is the overall order for the reaction? c. Calculate the rate constant? d. What is the rate for experiment 4? a. Looking at expts 1 and 2 the conc of A is constant = x 25C. What is the = x = 1 Looking at expts 1 and 3 the conc. of B is constant

y = = y = 2 rate law is rate=[NO]A]2[NO]B]1 Ch 15 Answer Key 3. continued b. 3 c. since rate=[NO]A]2[NO]B]1 using data from expt 1 0.141M/s=k(0.0019M)2(0.031M) k = 1.26x106 M-2s-1 d. since rate=[NO]A]2[NO]B]1 rate=(1.26x106 M-2s-1)(0.65M) 2(0.24M)= 1.28x105M/s Ch 15 Answer Key 4.

Given the following data for the reaction: 2A + B 2C Determine the rate law, the rate constant and the missing concentrations. Using expts 1 and 2 x = x=2 Using expts 2 and 3 y = y=0 So the rate law rate = k[NO]B]2 Using expt 1 0.00084M/min = k(0.72M) 2 k = 0.00162M-1min-1 For expt 4 0.0046M/min = (0.00162M-1min-1)[NO]B]2 [NO]B] = 1.68M

Ch 15 Answer Key 5. The following data was collected for the reaction A products. a. What is the rate law for the reaction? since a plot of ln[NO]A] vs. time gave a straight line it tells us that it is 1st order with respect to A rate = k[NO]A] b. What is the rate constant? k = slope slope = = = - 0.1s-1 k = 0.1 s-1 c. What is the concentration of A after 31 sec? the integrated rate law for 1st order is ln[NO]A] = kt + ln[NO]A]0 from the plot we can see that ln[NO]A]0 = -2.4 ln[NO]A] = -(0.1 s-1)(31s) + (-2.4) [NO]A] = 0.7M

Ch 15 Answer Key 6. Consider the decomposition of nitrogen dioxide: 2 NO2 2 NO + O2 Rate = k[NO2]2 A plot of 1/[NO2] verses time yielded a straight line with a slope of 1.8 x 10-3 Lmol-1s-1 at 500 K. If the initial concentration is 1.2 M, how long will it take for the [NO2] to decrease by 38%? 2nd order for 2nd order k = slope [NO]A] = 1.2M and if A has decreased by 38% theres 62% left so [NO]A]0 = (1.2M)(0.62) = 0.744M

t = 284s Ch 15 Answer Key 7. 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 680C. a. Calculate the time required for 95% of the phosphine to decompose. for 1st order reactions t1/2 =0.693/k k = 0.693/35s k = 0.0198s1 if 95% has decomposed theres 5% left ln(5) = -(0.0198)(t) + ln(100) t = 151s NOTE %s can only be used in 1st order rate laws b. What percentage of phosphine reacted after 78 seconds? ln[NO]A] = -(0.0198)(78s) + ln(100) [NO]A] = 21.3 21.3% remains so

78.7% has reacted Ch 15 Answer Key 8. Consider the following elementary (single-step) reaction: 2 NO2 (g) 2 NO (g) + O2 (g) The rate constant (k) for this reaction is 0.86 M-1 s -1 at 600K. Into a rigid, evacuated container, 2.50 atm of NO2 is added. Calculate the total pressure in the container at 600K after 20 seconds. since this is an elementary step the coefficient tells you its 2nd order with respect to NO2 = kt + because we want pressure we need to

change the unit of k PV=nRT M-1= = k = 0.86 M-1 s -1 x = 16.94 atm-1s-1 = (16.94 atm-1s-1)(20s) + P = 0.0029 atm continue to next slide Ch 15 Answer Key 8. continued 2 NO2 (g) initial change final

2 NO (g) O2 (g) 2.5 atm 0 0 - 2.4571 +2.4571 +1/2(2.4571) 0.0029 atm +2.4571 atm +1.2486 atm Ptotal = 0.0029 atm + 2.4571 atm + 1.2486 atm = 3.75 atm

Ch 15 Answer Key 9. Based on the following data determine the order for the reactant: If the half lives get longer by doubling 2nd order If the half lives get shorter by halving zero order But in this case the half lives are staying constant 1 st order Ch 15 Answer Key 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? Since the half-life times is getting shorter zero order

Use the half-life to get k t = 20min = k = 0.0025min-1M-1 Use the zero order integrated rate law [NO]A]=-kt+[NO]A] 0 [NO]A]=-(0.0025min-1M-1)(38min) + (0.1M) [NO]A]=0.005M Ch 15 Answer Key 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 ka = 4.5 x 10-4 s-1 and kb = 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 1st order ln[NO]A]=-kat + ln[NO]A]0 and ln[NO]B] = -kbt + ln[NO]B]0

since they have the same initial concentrations ln[NO]A] + kat = ln[NO]B] + kbt since [NO]A]=4[NO]B] ln4[NO]B] + kat = ln[NO]B] + kbt ln4 = kbt kat ln4 = t(3.7 x 10-3 s-1 - 4.5 x 10-4 s-1 ) t = 427s Ch 15 Answer Key 12. Consider the following reaction in which HCl adds across the double bond of ethylene: HCl + H2C=CH2 H3CCH2Cl The following mechanism, with the energy diagram shown below, has been suggested for this reaction: Step 1: HCl + H2C=CH2 H3CCH2 + + Cl Step 2: H3CCH2 + + Cl H3CCH2Cl

a. The rate limiting step is the slowest step and therefore will have greatest activation energy therefore step 1 is the rate limiting step b. The rate law can be derived from the slowest step of a mechanism where the coefficients are the orders rate = k[NO]HCl][NO]H 2C=CH2] the since it is first order with respect to both reactants the overall order is 2 c. The high energy loci on the graph (2 hill tops) are the transition states Ch 15 Answer Key 13. Given the following mechanism:

H2O2 H2O + O O + CF2Cl2 ClO + CF2Cl ClO + O3 Cl + 2 O2 Cl + CF2Cl CF2Cl2 a. Write the overall equation for the reaction? sum of the elementary steps if you add them up the O, CF2Cl2, ClO, CF2Cl and Cl cancel out leaving you with H2O2 + O3 H2O + 2 O2 b. Identify the reaction intermediates. O, ClO, CF2Cl and Cl reaction intermediates are temporarily made then consumed c. Identify the catalyst. CF2Cl2 catalysts speed up the reaction by changing the pathway and lowering the activation energy catalysts are temporarily consumed then regenerated

Ch 15 Answer Key 14. Write the overall reaction, identify any intermediates and derive a rate law given the following reaction mechanism: Cl2 2 Cl (fast equilibrium) Cl + CHCl3 HCl + CCl3 (slow) Cl + CCl3 CCl4 (fast) the rate law can be derived from the slowest elementary step rate = k2[NO]Cl][NO]CHCl3] the coefficient=order ONLY for elementary steps the rate law is not

practical if it contains intermediates such as in this case the Cl in step 1 it is a fast equilibrium so the rate forward=rate reverse k1[NO]Cl2] = k-1[NO]Cl]2 [NO]Cl] = now substituting this into the original rate law rate = k2[NO]CHCl3] or rate = k[NO]Cl2] [NO]CHCl3] Ch 15 Answer Key 15. Derive a rate law given the following reaction mechanism: Cl2 2 Cl (fast equilibrium) Cl + CO COCl (fast equilibrium) COCl + Cl2 COCl2 + Cl

(Slow) 2Cl Cl2 (Fast) Step 3 is the slow step rate = k3[NO]COCl][NO]Cl2] since COCl is an intermediate k2[NO]Cl][NO]CO] = k-2[NO]COCl] [NO]COCl] = substitute this into the rate law rate = k3[NO]Cl2] since Cl is an intermediate k1[NO]Cl2] = k-1[NO]Cl]2 [NO]Cl] = now substituting this into the original rate law rate = 1/2 [NO]CO][NO]Cl2] 3/2 Ch 15 Answer Key 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. E= forward EE= forward Ea reverse Ea E= forward EE= 42kJ/mol 32kJ/mol = 10kJ/mol Ch 15 Answer Key 17. The half-life for a first order reaction is 28 min at 300K. What is the half-life at 600K if the activation energy is 34 kJ/mol? If the T is increased the rate of the reaction increases due to a higher fractions of collisions with sufficient energy ln = since k = ln = ln =

ln = half life at 600 K is 0.031 min Ch 15 Answer Key 18. The activation energy for the reaction H2 (g) + I2 (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. k=Ae-Ea/RT = = = (59kJ/mol)/(0.0083145kJ/molK x 600K) = 11.827

= (184kJ/mol)/(0.0083145kJ/molK x 600K) = 36.883 = = 7.6x1010