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?

Answer 1

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

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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")`