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HOW TO FIND A RATE LAW

HOW TO FIND A RATE LAW. I. To find a “Rate Law,” use a series of experimental results. Hold the concentration of one reagent constant, vary the concentration of the other and note what happens to the rate. For example, for the reaction : 2 NO (g) + Cl 2 (g)  2 NOCl (g)

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HOW TO FIND A RATE LAW

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  1. HOW TO FIND A RATE LAW

  2. I. To find a “Rate Law,” use a series of experimental results. Hold the concentration of one reagent constant, vary the concentration of the other and note what happens to the rate. For example, for the reaction : 2 NO (g) + Cl2 (g)  2 NOCl (g) Experiment [NO]0 [Cl2]0 Rate, initial, M/sec 1. 0.10 0.10 3.0 x 10-5 2. 0.10 0.20 6.0 x 10-5 3. 0.20 0.10 1.2 x 10-4 The rate is expressed in general terms as : Rate = k [NO]x [Cl2]y and you need to determine x and y

  3. A. To determine the value of the exponents, compare the concentrations of one reagent in two experiments. The concentration of any other reagent must be kept constant. ([A]2 /[A]1) y = (Rate2/Rate1) where y is the exponent in the rate !

  4. B. Using Experiments 1 and 2 (where [NO]0 is constant) ([Cl2]2 /[Cl2]1) y = ([0.20]2/[0.10]1) = (6.0 x 10-5 M/sec/3.0 x 10-5 M/sec ) so ... 2y = 2 and y = 1 the exponent for [Cl2] C. Using Experiments 1 and 3 (where [Cl2]0 is constant) ([NO]3 /[NO]1) x = ([0.20]3/[0.10]1) = (1.2 x 10-4 M/sec/3.0 x 10-5 M/sec ) so ... 2x = 4 and x = 2 the exponent for [NO]

  5. D. The rate constant, k, can now be calculated. Use data from any one experiment : k = Rate/[NO]2 [Cl2] For experiment #1: k = 3.0 x 10-5 M/sec/(0.10 M)2(0.10 M) k = 3.0 x 10-2 M-2/sec For experiment #2: k = 6.0 x 10-5 M/sec/(0.10 M)2(0.20 M) k = 3.0 x 10-2 M-2/sec

  6. II. Examples : A. NO2(g) + CO(g)  CO2 (g) + NO (g) Experiment [NO2]0 [CO]0 Rate, initial, M/sec 1. 0.10 0.10 1.0 2. 0.10 0.020 1.0 3. 0.10 0.010 1.0 4. 0.020 0.10 0.040 5. 0.010 0.10 0.010

  7. Using Experiments 1 and 2, as [CO] doubles, Rate remains constant ! So far, rate = k [CO]0 ! From Experiments 5 and 4, [NO2] doubles, Rate quadruples ! So now, Rate = k [NO2]2 [CO]0 or Rate = k [NO2]2 ! This Rate Law is second order in NO2 and zero order in CO !

  8. B. Practice on this one : 3 A + B + 2 C  3D + E Rate Experiment initial [A]0 [B]0 [C]0 M/sec 1. 0.10 0.10 0.20 2.0 2. 0.30 0.10 0.20 18.0 3. 0.30 0.20 0.20 18.0 4. 0.10 0.10 0.40 16.0 Determine the rate law ? Calculate the rate constant ?

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