# Question c81c4

Nov 3, 2015

$t = 248791.2 \sec = 69.1 t$

#### Explanation:

$r a t e = - \frac{\mathrm{dA}}{\mathrm{dt}} = k {\left[A\right]}^{2}$

Integrated rate Law:
$\frac{1}{{\left[A\right]}_{t}} = k \cdot t + \frac{1}{{\left[A\right]}_{0}}$

0,249 M of A is used = 95.77%
0.011 M of A is not used = 4.23%#

fraction of A not used: $\frac{4.23}{100} = \frac{1}{23.64}$

${\left[A\right]}_{t} = \frac{{\left[A\right]}_{0}}{23.64}$

and $t = {t}_{\frac{1}{23.64}}$

$\frac{1}{\frac{{\left[A\right]}_{0}}{23.64}} = k \cdot {t}_{\frac{1}{23.64}} + \frac{1}{A} _ 0$

${t}_{\frac{1}{23.64}} = \frac{1}{k} \left(\frac{23.64}{A} _ 0 - \frac{1}{A} _ 0\right) = \frac{1}{3.5 \cdot {10}^{- 4}} \left(\frac{22.64}{0.260}\right)$

${t}_{\frac{1}{23.64}} = 248791.2 \sec = 69.1 t$