Answer:
Option D, Concentration of NO2 decomposes after 4.00 s = 0.77 mol
Explanation:

Time (t) = 4.00\;s
Initial concentration of NO2 = 1.33 M
Integrated law for second order reaction:
![\frac{1}{[A]}=\frac{1}{[A]_0} =kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%3Dkt)
Where, [A] = Concentration after time, t
[A]0 = Intitial concentration, k = rate constant, t = time
On substituting values in the above
![\frac{1}{[A]}=\frac{1}{1.33} =0.255 \times 4.00](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B1.33%7D%20%3D0.255%20%5Ctimes%204.00)
![\frac{1}{[A]} =1.772](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D1.772)
[A] = 0.5644 M
Concentration of NO2 decomposes after 4.00 s = 1.33 - 0.5644 = 0.7656 M
No. of mole = Molarity * volume
= 0.7656 * 1
= 0.7656 mol 0r 0.77 mol