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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 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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slide2
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

slide3
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 !

slide4
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]

slide5
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

slide6
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

slide7
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 !

slide8
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 ?