It'll take **0.061 minutes** for the concentration of nitrogen dioxide to reach **0.28 M**.

#2NO_(2(g)) -> 2NO_((g)) + O_(2(g))#

The integrated rate law for a *second-order* reaction looks like this

#1/([A]) - 1/([A_0]) = k * t#, where

#[A_0]# - the initial concentration of nitrogen dioxide;

#[A]# - the concentration of nitrogn dioxide after the passing of a certain amount of time, #t#;

#k# - the rate constant - in your case, #k = "0.54 M"^(-1)"s"^(-1)#

#t# - time;

Since you've got all you need to solve for #t#, plug your values into the above equation

#1/([A]) - 1/([A_0]) = k * t => t= (1/([A]) - 1/([A_0]))/k#

#t= (1/([A]) - 1/([A_0])) * 1/k = (1/0.28 - 1/0.62)cancel("M"^(-1)) * 1/(0.54cancel("M"^(-1))"s"^(-1)#

#t = 1.9585/("0.54 s"^(-1)) = "3.627 s"#

Since you want the time expressed in minutes, the answer will be (rounded to two sig figs)

#3.627cancel("s") * "1 min"/(60cancel("s")) = "0.0605 min"#

#t = color(green)("0.061 min")#