Integrated rate law half life sample problems

Question

The rate of the reaction O + NO₂ ----------> NO + O_{2 was studied at a certain temperature.}

time	O (atoms/cm3)
0	5.0 x 109
1.0 x 10-2	1.9 x 109
2.0 x 10-2	6.8 x 108
3.0 x 10-2	2.5 x 108

1) In the first set of experiments, NO₂ was in large excess at a concentration of 1.0 x 10¹³ molecules/cm³with the following data collected above in the table.

1a) what is the order of the reaction with respect to oxygen atoms?

2) The reaction is known to be first order with respect to NO_{2 .} determine the overall rate law and the law and the value of the rate constant?

Answer 1

Since NO₂ was in large excess, it can be ignored in determining the order with respect to oxygen. We will find the rate of disappearance of O with respect to time (slope) to determine the order. You could graph the data, but that's probably not what you are looking for. So, here we go...

slope 1 = 5.0x10⁹ - 1.9x10⁹ / 1.0x10^-2 = 3.1x10⁹ / 1.0x10^-2 = 3.1x10¹¹

slope 2 = 1.9x10⁹ - 0.68x10⁹ / 1x10^-2 = 1.2x1011 (note: slopes are not the same so is NOT zero order)

Redo using 1/oxygen vs time:

1/5.0x109 = 2x10-10

1/1.9x109 = 5.3x10-10

1/0.68x109 = 14.7x10-10

slope 1 = 5.3 - 2 / 1 = 3.3

slope 2 = 14.7 - 5.3 /1 = 9.4 (note: slopes are not the same so is NOT 2nd order)

Using ln (natural log) oxygen vs time

ln 5x109 = 22.3

ln 1.9x109 = 21.4

ln 0.68x109 = 20.3

slope 1 = 22.3 - 21.4 /1 = 0.9

slope 2 = 21.4 - 20.3 = 1.1 (note: Slopes are virtually the same indicated FIRST ORDER)

Overall rate law: rate = k[O][NO₂]

Rate constant: k = rate / [O][NO₂] = 3.1x10¹¹ / [5x10⁹][1x10¹³] = answer