JEE Main 2025 · Chemistry · Chemical Equilibrium · medium

Gaseous N2O4 at a total pressure of 2 atm and a fixed temperature has an equilibrium vapour density of 34.5 (the vapour density of undissociated N2O4…

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Gaseous

\mathrm{N_2O_4}

at a total pressure of

2\ \mathrm{atm}

and a fixed temperature has an equilibrium vapour density of

34.5

(the vapour density of undissociated

\mathrm{N_2O_4}

46

) for

\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)}

. At the same temperature, the total pressure at which the degree of dissociation falls to

0.2

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Answer: (D) $6\ \mathrm{atm}$

Initial vapour density

D

and observed

d

satisfy

D/d = 1+\alpha

, so

1+\alpha = 46/34.5 = 4/3

and

\alpha=1/3

. For this equilibrium

K_p=\dfrac{4\alpha^2P}{1-\alpha^2}=\dfrac{4(1/9)(2)}{8/9}=1

atm. At

\alpha=0.2

P=\dfrac{K_p(1-\alpha^2)}{4\alpha^2}=\dfrac{0.96}{0.16}=6\ \mathrm{atm}

. Increasing pressure suppresses dissociation, consistent with the larger value.

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