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HOW TO FIND A RATE LAW

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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 ... – PowerPoint PPT presentation

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Title: HOW TO FIND A RATE LAW


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 NO0 Cl20 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 NOx Cl2y
  • 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.
  • (A2 /A1) y (Rate2/Rate1)
  • where y is the exponent in the rate !

4
  • B. Using Experiments 1 and 2 (where NO0 is
    constant)
  • (Cl22 /Cl21) y (0.202/0.101)
  • (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 Cl20 is
    constant)
  • (NO3 /NO1) x (0.203/0.101)
  • (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/NO2 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 NO20 CO0 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 CO0 !
  • From Experiments 5 and 4,
  • NO2 doubles,
  • Rate quadruples !
  • So now, Rate k NO22 CO0
  • or Rate k NO22 !
  • 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
  • A0 B0 C0 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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