# 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?

Jan 29, 2017

I got $\text{71.4 s}$, after we have gone through the two half-lives for ${\text{NO}}_{2}$.

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

$2 {\text{NO"_2(g) -> 2"NO"(g) + "O}}_{2} \left(g\right)$,

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 $r \left(t\right) = \pm \frac{1}{\nu} _ i \frac{\Delta \left[i\right]}{\Delta t}$ 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 $\Delta \left[{\text{NO}}_{2}\right]$ using infinitesimally small intervals, we notate it as $d \left[{\text{NO}}_{2}\right]$, and if we use infinitesimally small $\Delta t$, we notate it as $\mathrm{dt}$:

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 ${\left[{\text{NO}}_{2}\right]}_{0}$ at ${t}_{0} = \text{0 s}$, and our final state is $\left[{\text{NO}}_{2}\right]$ at $t = t$ $\text{s}$.

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

$2 k t = \frac{1}{{\left[{\text{NO"_2]) - 1/(["NO}}_{2}\right]}_{0}}$

For a half-life, the current concentration after ${t}_{\text{1/2}}$ time (i.e. one half-life) is $\frac{1}{2} {\left[{\text{NO}}_{2}\right]}_{0}$, so:

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

$= \frac{1}{{\left[{\text{NO}}_{2}\right]}_{0}}$

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 $2 k$.)

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

We know ${\left[{\text{NO}}_{2}\right]}_{0}$, we can solve for the amount of time it takes to get to $\frac{{\left[{\text{NO}}_{2}\right]}_{0}}{4}$.

Note that if we got to $\frac{1}{4}$ the initial concentration, we passed through two half-lives. Therefore:

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

$= \frac{1}{\left(1.40 \times {10}^{- 2} \text{M"^(-1)cdot"s"^(-1))("1.00 M}\right)}$

$=$ $\textcolor{b l u e}{\text{71.4 s}}$