$\mathrm{NO}_{2}$ required for a reaction is produced by the decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}$ in $\mathrm{CCl}_{4}$ as per the equation $2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ The initial concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is $3.00 \mathrm{~mol} \mathrm{~L}^{-1}$ and it is $2.75 \mathrm{~mol} \mathrm{~L}^{-1}$ after 30 minutes. The rate of formation of $\mathrm{NO}_{2}$ is:

(A) $1.667 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

(B) $4.167 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

(C) $8.333 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

(D) $2.083 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Chemical Kinetics

NO_{2}

required for a reaction is produced by the decomposition of

N_{2} O_{5}

CCl_{4}

as per the equation

2 N_{2} O_{5} (g) \to 4 NO_{2} (g) + O_{2} (g)

The initial concentration of

N_{2} O_{5}

3.00 mol L^{- 1}

and it is

2.75 mol L^{- 1}

after 30 minutes.
The rate of formation of

NO_{2}

is:

Answer: (A)

The disappearance of

N X_{2} O X_{5}

- \frac{Δ [ N X _{2} O X _{5} ]}{Δ t} = \frac{3.00 - 2.75}{30} = 8.33 \times 1 0^{- 3}

mol L

^{- 1}

min

^{- 1}

. From the stoichiometry

2 N X_{2} O X_{5} - > 4 NO X_{2} + O X_{2}

NO X_{2}

appears twice as fast as

N X_{2} O X_{5}

vanishes, so

\frac{d [ NO X _{2} ]}{d t} = 2 \times 8.33 \times 1 0^{- 3} = 1.667 \times 1 0^{- 2}

mol L

^{- 1}

min

^{- 1}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option