\(\mathrm{NO}_{2}\) required for a reaction is produced by decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) as by 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 \(3 \mathrm{~mol} \mathrm{~L}^{-1}\) and it is \(2.75 \mathrm{~mol} \mathrm{~L}^{-1}\) after 30 minutes. The rate of formation of \(\mathrm{NO}_{2}\) is \(\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \min ^{-1}\), value of \(x\) is ____.

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Chemical Kinetics

NO_{2}

required for a reaction is produced by decomposition of

N_{2} O_{5}

CCl_{4}

as by equation

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

The initial concentration of

N_{2} O_{5}

3 mol L^{- 1}

and it is

2.75 mol L^{- 1}

after 30 minutes.
The rate of formation of

NO_{2}

x \times 1 0^{- 3} mol L^{- 1} min^{- 1}

, value of

x

is ____.

Answer: 17

For

2 N_{2} O_{5} \to 4 NO_{2} + O_{2}

, the rate of reaction is related to the rate of change of each species through its stoichiometric coefficient:

- \frac{1}{2} \frac{Δ [ N _{2} O _{5} ]}{Δ t} = \frac{1}{4} \frac{Δ [ NO _{2} ]}{Δ t}

. With

[N_{2} O_{5}]

falling from 3 to 2.75 mol/L over 30 minutes,

- \frac{Δ [ N _{2} O _{5} ]}{Δ t} = \frac{0.25}{30} mol L^{- 1} min^{- 1}

, so the rate of reaction is

\frac{1}{2} \times \frac{0.25}{30} = \frac{1}{240} mol L^{- 1} min^{- 1}

. The rate of formation of

NO_{2}

is four times this:

\frac{4}{240} = 16.7 \times 1 0^{- 3} mol L^{- 1} min^{- 1} \approx 17 \times 1 0^{- 3} mol L^{- 1} min^{- 1}

, so

x = 17

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer