030 020 n 1 15 n log1 n log15 log1 log15 n Rate 4 Rate 1 C 2 C 3 p 72 x 10 6 24

# 030 020 n 1 15 n log1 n log15 log1 log15 n rate 4

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0.30 0.20 n 1 1.5 n log1 n log1.5 log1 log1.5 n 0 Rate 4 Rate 1 C 2 C 3 p 7.2 x 10 6 2.4 x 10 6 0.60 0.20 p 3 3 p log3 p log3 log3 log3 p 1 CHEM 111 Week 8 - Kinetics Fall 2014 The following data was obtained using the method of initial rates at at 25 o C: Experiment [NO 2 ] M [O 3 ] M Rate M s -1 1 0.21 0.70 6.30 10 -3 2 0.63 0.70 1.89 10 -2 3 0.21 0.35 3.15 10 -3 4 0.42 0.35 6.30 10 -3 a) Use the data shown to determine the rate law for the reaction. b) Using the results of any one of these experiments calculate the rate constant for this reaction. c) If the frequency factor (A) is 1 M -1 s -1 for this reaction what is the activation energy? d) It has been suggested that this reaction is an elementary reaction. Does the rate law you determined in part (a) support this assertion? Explain your answer. Solutions: a) Rate = k[NO 2 ][O 3 ] R 1 /R 2 = ([NO 2 ] 1 /[NO 2 ] 2 ) a 6.30x10 -3 M/s / 1.89x10 -2 M/s = (.21 M / .63 M) a .333 = .333 a , a=1 R 1 /R 3 = ([O 3 ] 1 /[O 3 ] 3 ) b 6.30x10 -3 M/s / 3.15x10 -3 M/s = (.70 M / .35 M) b 2 = 2 b , b=1 rate = k[NO 2 ] [O 3 ] b) 6.30 x10 -3 M/s = k [.42 M][.35 M] k = 4.3x10 -2 1/Ms c) k = Ae -Ea/RT ln (k/A) = -E a /RT E a = -RT ln (k/A) E a = -(.008315 kJ/mol K)(298 K) ln (4.3x10 -2 ) E a = 7.8 kJ/mol d) Yes 4) The reaction CHCl 3 ( g ) + Cl 2 ( g ) →CCl 4 ( g ) + HCl( g ) has the following rate law: Rate = k [CHCl 3 ][Cl 2 ]. If the concentration of CHCl3 is increased by a factor of five while the concentration of Cl2 is kept the same, the rate will? CHEM 111 Week 8 - Kinetics Fall 2014 Multiplied by 5.  #### You've reached the end of your free preview.

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• Spring '08
• Kenney
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