Explanation

(b)
For the first order reaction,
$2N_{2}O_{5} \rightarrow 2NO_{2} + O_{2}$
$rate\ \frac{dx}{dt} = k\left( N_{2}O_{5} \right)\ \ \ \ldots(i)$
Given,
$\frac{dx}{dt} = 2.400 \times 10^{- 5}mol\ L^{- 1}s^{- 1}$
$k = 3.0 \times 10^{- 5}s^{- 1}$
$\left\lbrack N_{2}O_{5} \right\rbrack = ?$
Substituting these values in (i)
$2.4 \times 10^{- 5} = 3.0 \times 10^{- 5}\left\lbrack N_{2}O_{5} \right\rbrack$
$or\ \left\lbrack N_{2}O_{5} \right\rbrack = \frac{2.4 \times 10^{- 5}}{3.0 \times 10^{- 5}}$
$= 0.8mol\ L^{- 1}$