The second-order decomposition of no2 has a rate constant of 0.255 m-15-1. how much no2 decomposes in 4.00 s if the initial concentration of no2 (1.00 l volume) is 1.33 m? a) 1.8 mol b) 0.85 mol c) 0.48 mol d) 0.77 mol

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

idk what is it? is this a question that you know the answer or u dont