# Question 161ec

Apr 13, 2015

It'll take 0.061 minutes for the concentration of nitrogen dioxide to reach 0.28 M.

$2 N {O}_{2 \left(g\right)} \to 2 N {O}_{\left(g\right)} + {O}_{2 \left(g\right)}$

The integrated rate law for a second-order reaction looks like this

$\frac{1}{\left[A\right]} - \frac{1}{\left[{A}_{0}\right]} = k \cdot t$, where

$\left[{A}_{0}\right]$ - the initial concentration of nitrogen dioxide;
$\left[A\right]$ - the concentration of nitrogn dioxide after the passing of a certain amount of time, $t$;
$k$ - the rate constant - in your case, $k = {\text{0.54 M"^(-1)"s}}^{- 1}$
$t$ - time;

Since you've got all you need to solve for $t$, plug your values into the above equation

$\frac{1}{\left[A\right]} - \frac{1}{\left[{A}_{0}\right]} = k \cdot t \implies t = \frac{\frac{1}{\left[A\right]} - \frac{1}{\left[{A}_{0}\right]}}{k}$

t= (1/([A]) - 1/([A_0])) * 1/k = (1/0.28 - 1/0.62)cancel("M"^(-1)) * 1/(0.54cancel("M"^(-1))"s"^(-1)

t = 1.9585/("0.54 s"^(-1)) = "3.627 s"

Since you want the time expressed in minutes, the answer will be (rounded to two sig figs)

3.627cancel("s") * "1 min"/(60cancel("s")) = "0.0605 min"#

$t = \textcolor{g r e e n}{\text{0.061 min}}$