J.R. S. answered • 11/08/19

Ph.D. University Professor with 10+ years Tutoring Experience

Since NO_{2} 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 = 5.0x10^{9} - 1.9x10^{9 }/ 1.0x10^{-2} = 3.1x10^{9 }/ 1.0x10^{-2} = 3.1x10^{11}

slope 2 = 1.9x10^{9} - 0.68x10^{9} / 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 1st 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 **SECOND ORDER**)

Overall rate law: rate = k[O]^{2}[NO_{2}]

Rate constant: k = rate / [O]^{2}[NO_{2}] = 3.1x10^{11} / [5x10^{9}]^{2}[1x10^{13}] = 3.1x10^{11} / 2.5x10^{30}

k = 1.24x10^{-19} cm3^{2} atoms-^{2} time^{-1}