Rate Law from Concentration versus Time Data

NO2(g) + CO(g) NO(g) + CO2(g)

The rate of the above reaction depends only on the concentration of nitrogen dioxide at temperatures below 225°C.
At a temperature below 225°C the following data were obtained:

Time (s) [NO2] (mol/L)
0 0.420
9.00×102 0.403
2.25×103 0.380
3.38×103 0.362
6.75×103 0.319
1.35×104 0.257

I found k[NO2]2 = -d[NO2]/dt & 1/[NO2] - 1/[NO2]0 = kt

Calculate the rate constant (k).
Enter the numerical value and corresponding units.

you know this..
-d[A] / dt = k x [NO2]^n

right? but you don't know what n is do you?

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the integrated rate law equations are..

[A] = -kt + [Ao].. ... ... for n=0
ln[A] = -kt + ln[Ao].... for n=1
1/[A] = +kt + 1/[Ao]... for n=2

and since k and [Ao] are constant, all of those are of the form y = mx + b if we let y = [A] or ln[A] or 1/[A] (depending on whether n = 0, 1, or 2) and x = t

ie..
if n=0... a plot of [A] vs t will give a straight line with slope = -k and intercept = [Ao]
if n=1.. ln[A] vs t will give a line with slope = -k and intercept = ln[Ao]
if n=2..1/[A] vs t will give a line with slope = +k and intercept = 1/[Ao]

so.. let's make some plots and find out whether this is zero, first or 2nd order.

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open up an excel spreadsheet and make this table
... ... . .... .col A... ... col B.... .. .col C... .... col D
row 1.. .. .time..... [NO2]..... ln[NO2]... .1/[NO2]
row 2... .. .0.. ... .. ..0.420... .. =ln(B2)... ..=1/B2
row 3... .. 900.. ... ..0.403... .. =ln(B3)... ..=1/B3
etc..
row 7... .13500... .. 0.257.. ... .. etc..

and the "=" makes the cell a formula.. and you can copy it down through row 7 so that you get something like this

... ... . .... .col A... ... col B.... .. .col C... .... col D
row 1.. .. .time..... [NO2]..... ln[NO2]... .1/[NO2]
row 2... .. .0.. ... .. ..0.420... ..-0.8675... .2.3809