Question

At 573 K, gaseous NO2(g) decomposes, forming NO(g)
and O2(g). If a vessel containing NO2(g) has an initial
concentration of 1.9 × 10−2 mol/L, how long will it take
for 75% of the NO2(g) to decompose? The decomposition
of NO2(g) is second-order in the reactant and the this reaction, at 573 K, is 1.1 L/mol ∙ s.

All I know is that the second order rate equation is 1/[At]=1/[Ao]+kt
Any point in the right direction would be great! I'm really lost on this. The answer is 140 seconds, but how do I get that?

Answer 1

A_t = 0.019 x 0.25 = ? (If 75% decomposes it must leave 25%).
A_o = 0.019
You know k.
Plug into the equation and solve for t. I get 143.5 s and that rounds to 140 s to 2 s.f. (dictated by 0.019)

Answer 2

97 s