Answer
(a) $[N_2~O_5](t) = C~e^{-0.0005~t}$
(b) It takes $~~211~s~~$ to reach 90% of the original concentration.
Work Step by Step
(a) As the reaction continues, the concentration of $[N_2~O_5]$ will decrease at this rate:
$\frac{d[N_2~O_5]}{dt} = -0.0005~[N_2~O_5]$
Then:
$[N_2~O_5](t) = C~e^{-0.0005~t}$
(b) We can find the time it takes to reach 90% of the original concentration:
$[N_2~O_5](t) = C~e^{-0.0005~t} = 0.90~C$
$C~e^{-0.0005~t} = 0.90~C$
$e^{-0.0005~t} = 0.90$
$-0.0005~t = ln(0.90)$
$t = \frac{ln(0.90)}{-0.0005}$
$t = 211~s$
It takes $~~211~s~~$ to reach 90% of the original concentration.
