For the reaction 2 ~N_2 O_5_(8) arrow 4 NO_2_(g)+O_2_(8) initial concentration of N_2 O_5 is 2.0 molL^-1 and after 300 min , it is reduced to 1.4 molL^-1 . The rate of production of NO_2 ( . in molL^-1 ~min^-1 ) is

Correct answer: (D) 4 × 10^-3. Rate of formation of NO_2=2 × rate of disappearance of N_2 O_5 =2 × [(-(1.4-2))/300 ] =4 × 10^-3

Q56 / 180 · KCET 2025 · Chemistry

For the reaction $2 \mathrm{~N}_2 \mathrm{O}_{5_{(8)}} \rightarrow 4 \mathrm{NO}_{2_{(\mathrm{g})}}+\mathrm{O}_{2_{(8)}}$ initial concentration of $\mathrm{N}_2 \mathrm{O}_5$ is $2.0 \mathrm{molL}^{-1}$ and after 300 $\min$, it is reduced to $1.4 \mathrm{molL}^{-1}$. The rate of production of $\mathrm{NO}_2\left(\right.$ in $\mathrm{molL}^{-1} \mathrm{~min}^{-1}$ ) is

A
$2.5 \times 10^{-4}$
B
$4 \times 10^{-4}$
C
$2.5 \times 10^{-3}$
Correct answer
D
$4 \times 10^{-3}$

Solution

Rate of formation of $\mathrm{NO}_2=2 \times$ rate of disappearance of $\mathrm{N}_2 \mathrm{O}_5$ $\begin{aligned} & =2 \times\left[\frac{-(1.4-2)}{300}\right] \\ & =4 \times 10^{-3}\end{aligned}$

Hence, the correct answer is option: 'D'.

Solution

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