\(37.8 \mathrm{~g} \mathrm{~N}_{2} \mathrm{O}_{5}\) was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K \(2 \mathrm{~N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\) The total pressure at equilibrium was found to be 18.65 bar. Then, \(\mathrm{Kp}=\) \(\_\_\_\_\) \(\times 10^{-2}\) [nearest integer] Assume \(\mathrm{N}_{2} \mathrm{O}_{5}\) to behave ideally under these conditions Given : \(\mathrm{R}=0.082\) bar L mol \(^{-1} \mathrm{~K}^{-1}\)

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Equilibrium

37.8 g N_{2} O_{5}

was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K

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

The total pressure at equilibrium was found to be 18.65 bar.

Then,

Kp =

____

\times 1 0^{- 2}

[nearest integer] Assume

N_{2} O_{5}

to behave ideally under these conditions
Given :

R = 0.082

bar L mol

^{- 1} K^{- 1}

Answer: 962

Moles of

N_{2} O_{5}

: molar mass

= 108 g / m o l

, so

n = \frac{37.8}{108} = 0.35 m o l

.

Initial pressure in the 1 L vessel at 500 K:

P_{0} = \frac{n R T}{V} = \frac{0.35 \times 0.082 \times 500}{1} = 14.35 ba r

For

2 N_{2} O_{5} ⇌ 2 N_{2} O_{4} + O_{2}

, let the pressure of

O_{2}

formed at equilibrium be

P

; then

N_{2} O_{4}

formed is

2 P

and

N_{2} O_{5}

consumed is

2 P

P_{N_{2} O_{5}} = 14.35 - 2 P, P_{N_{2} O_{4}} = 2 P, P_{O_{2}} = P

Total pressure:

14.35 - 2 P + 2 P + P = 14.35 + P = 18.65 \Rightarrow P = 4.3 ba r

P_{N_{2} O_{5}} = 5.75 ba r

P_{N_{2} O_{4}} = 8.6 ba r

P_{O_{2}} = 4.3 ba r

K_{p} = \frac{P _{N_{2} O_{4}}^{2} \cdot P _{O_{2}}}{P _{N_{2} O_{5}}^{2}} = \frac{( 8.6 ) ^{2} \times 4.3}{( 5.75 ) ^{2}} = 9.62

K_{p} = 962 \times 1 0^{- 2}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Integer answer