For a reaction, $\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$ in a constant volume container, no products were present initially. The final pressure of the system when $50 \%$ of reaction gets completed is

(D) $7 / 4$ times of initial pressure

(A) $7 / 2$ times of initial pressure

(B) 5 times of initial pressure

(C) $5 / 2$ times of initial pressure

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Chemical Kinetics

For a reaction,

N_{2} O_{5 (g)} \to 2 NO_{2 (g)} + \frac{1}{2} O_{2 (g)}

in a constant volume container, no products were present initially. The final pressure of the system when

50%

of reaction gets completed is

Answer: (D)

Let the initial pressure of pure

N_{2} O_{5}

P_{0}

. For

N_{2} O_{5} (g) \to 2 N O_{2} (g) + \frac{1}{2} O_{2} (g)

, when 50% of the

N_{2} O_{5}

has reacted:

N_{2} O_{5}

remaining

= 0.5 P_{0}

N O_{2}

formed

= 2 \times 0.5 P_{0} = P_{0}

O_{2}

formed

= \frac{1}{2} \times 0.5 P_{0} = 0.25 P_{0}

Total pressure

= 0.5 P_{0} + P_{0} + 0.25 P_{0} = 1.75 P_{0} = \frac{7}{4} P_{0}

The final pressure is

7/4

times the initial pressure.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option