Question

$NO_2$ required for a reaction is produced by the decomposition of $N_2O_5$ in $CCl_4$ as per the equation
$ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g).$
The initial concentration of $N_2O_5$ is $3.00\, mol$ $L^{-1}$ and it is $2.75\, mol$ $L^{-1}$ after $30$ minutes. The rate of formation of $NO_2$ is :

• A. $2.083 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}$
• B. $4.167 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}$
• C. $8.333 \times 10^{-3} \; mol \; L^{-1} \; min^{-1}$
• D. $1.667 \times 10^{-2} \; mol \; L^{-1} \; min^{-1}$

Answer 1

Answer: (D)

$ 2N_2 O_5 (g) \rightarrow 4NO_2(g) + O_2(g) $
$3.0 M$
$2.75 $ M
$\frac{-\Delta \left[N_{2}O_{5}\right]}{\Delta t} = \frac{0.25}{30} $
$ \frac{1}{2} \times\frac{-\Delta \left[N_{2}O_{5}\right]}{\Delta t} = \frac{1}{4} \times\frac{- \Delta \left[N_{2}O_{5}\right]}{\Delta t} = \frac{1}{4} \times\frac{\Delta \left[NO_{2}\right]}{\Delta t}$
$ \frac{\Delta \left[NO_{2}\right]}{\Delta t} = \frac{0.25}{30} \times2 = 1.66 \times10^{-2} M /\min$