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?
1 Answer
I got
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
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
#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
#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
#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
#color(green)(t_"1/2" = 1/(2k["NO"_2]_0))#
(Keep in mind that if the reactant had a stoichiometric coefficient of
We assume that the rate constant is
We know
Note that if we got to
#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")#