Dale Malull, 14 (65 Home Work Ch ue pts) Due Dalu Walllll 1. Relate the rate of

Question

Dale Malull, 14 (65 Home Work Ch ue pts) Due Dalu Walllll 1. Relate the rate of decomposition of No to the rate of formation of O2 for the following reaction: 8pts 2NO2 (g) 2NO(g) O2 (g) 2. Nitric oxide, NO, reacts with hydrogen to give nitrous oxide N20 and water. 2NO( g) H2 (g) N2 O (g) H2 O (g) 12 pts Initial Conc. NO Initial Conc H Initial Rate Expt 1 6.4 x 10-3 M 2.2x10-3 M 2.6x10-5 M/s Expt 2 12.8 x 10-3 M 2.2 x 10-3 M 1.0 x 104 M/s Expt 3 6.4 x 10-3 M 4.5 x 10-3 M 5.1 x 10 M/s Find out the rate law and the value of the rate constant for the reaction of NO. 3. A decomposes to give B and C. A--- B+ C 9 pts

Explanation / Answer

To determine the rate law, which is of the following form:

rate law = k[NO]a[H2]b

We need to obtain the exponents of the reactants, as follows:

For a:

Initial rate 1 / Initial rate 2 = k[NO]1a[H2]1b / k[NO]2a[H2]2b

2.6 x 10-5 / 1 x 10-4 = k[6.4 x 10-3]a[2.2 x 10-3]b / k[12.8 x 10-3]a[2.2 x 10-3]b

Cancelling out:

2.6 x 10-5 / 1 x 10-4 =  [6.4 x 10-3]a/ [12.8 x 10-3]a

0.26 = 0.5a

ln (0.26) = a ln (0.5)

a = 1.9434

For b:

Initial rate 1 / Initial rate 3 = k[NO]1a[H2]1b / k[NO]3a[H2]3b

2.6 x 10-5 / 5.1 x 10-5 = k[6.4 x 10-3]a[2.2 x 10-3]b / k[6.4 x 10-3]a[4.5 x 10-3]b

Cancelling out:

2.6 x 10-5 / 5.1 x 10-5 =  [2.2 x 10-3]b/ [4.5 x 10-3]b

0.5098 = 0.488889b

ln (0.5098) = b ln (0.488889)

b= 0.94146

So rate law will be:

rate law = k[NO]1.9434[H2]0.94146

To get the rate constant k, we substitute values of any row in the table given, as follows (using row 1):

2.6 x 10-5 = k[6.4x10-3]1.9434[2.2x10-3]0.94146

k = 151.51 M-1.88486 s-1

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