The reaction #2NO_2 -> 2NO + O_2# obeys the rate law #(Delta[O_2])/(Deltat) = (1.40 xx 10^-2 )[NO_2]^2# at 500 K. If the initial concentration of #NO_2# is 1.00 M, how long will it take for the #[NO_2]# to decrease to 25.0% of its initial value?

I got #"71.4 s"#, after we have gone through the two half-lives for #"NO"_2#.

This seems to be a second-order half-life (yes, that is a thing). I don't know the equation off the top of my head, but we can derive it.

For the reaction

#2"NO"_2(g) -> 2"NO"(g) + "O"_2(g)#,

we have the rate law

#r(t) = k["NO"_2]^2 = -1/2(Delta["NO"_2])/(Deltat) = 1/1(Delta["O"_2])/(Deltat)#,

once we recall that #r(t) = pm1/nu_i (Delta[i])/(Deltat)# describes the rate of disappearance of reactant or appearance of product, where #nu# is the stoichiometric coefficient.

To derive the second-order half-life equation:

#k["NO"_2]^2 = -1/2(Delta["NO"_2])/(Deltat)#

#2kDeltat = -1/(["NO"_2]^2)Delta["NO"_2]#

If we treat #Delta["NO"_2]# using infinitesimally small intervals, we notate it as #d["NO"_2]#, and if we use infinitesimally small #Deltat#, we notate it as #dt#:

#2kdt = -1/(["NO"_2]^2)d["NO"_2]#

Now if we add up all the infinitesimally small intervals over the range of the reaction, our initial state is #["NO"_2]_0# at #t_0 = "0 s"#, and our final state is #["NO"_2]# at #t = t# #"s"#.

#2int_(0)^(t) kdt = -int_(["NO"_2]_0)^(["NO"_2]) 1/(["NO"_2]^2)d["NO"_2]#

#2kt = 1/(["NO"_2]) - 1/(["NO"_2]_0)#

For a half-life, the current concentration after #t_"1/2"# time (i.e. one half-life) is #1/2["NO"_2]_0#, so:

#2kt_"1/2" = 2/(["NO"_2]_0) - 1/(["NO"_2]_0)#

#= 1/(["NO"_2]_0)#

This gives the second-order half-life, for a reactant with a stoichiometric coefficient of #2#, as:

#color(green)(t_"1/2" = 1/(2k["NO"_2]_0))#

(Keep in mind that if the reactant had a stoichiometric coefficient of #1#, let's say, then the denominator has #k#, not #2k#.)

We assume that the rate constant is #k = 1.40 xx 10^(-2)# #"M"^(1)cdot"s"^(-1)#, based on its magnitude in the rate law (the timescale appears to be on the order of seconds).

We know #["NO"_2]_0#, we can solve for the amount of time it takes to get to #(["NO"_2]_0)/4#.

Note that if we got to #1/4# the initial concentration, we passed through two half-lives. Therefore:

#color(blue)(t) = 2t_"1/2" = 1/(k["NO"_2]_0)#

#= 1/((1.40 xx 10^(-2) "M"^(-1)cdot"s"^(-1))("1.00 M"))#

#=# #color(blue)("71.4 s")#