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?

Question

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?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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

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

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:

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

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

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

The reaction $2NO_2 -&gt; 2NO + O_2$ 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 Answers

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?