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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:
Where, [A] = Concentration after time, t
[A]0 = Intitial concentration, k = rate constant, t = time
On substituting values in the above
[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