$K_{p} f o r N_{2} O_{2} ⇀ {2 N O}_{2}$ at equilibrium pressure of ${3 P}_{o}$ is $0.5 P_{o}$ . On halving the volume of container equilibrium is distributed. The new equilibrium pressure is :

[\frac{79 - \sqrt{161}}{16}] P_{o}

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[\frac{79 + \sqrt{161}}{16}] P_{o}

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[\frac{17 - \sqrt{161}}{16}] P_{o}

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Solution

The correct option is D $N o n e o f t h e s e$
$\begin{matrix} Reaction: & N_{2} O_{4} ⇌ 2 {N O}_{2} t & = 0 p 0 t & = t_{e q} (p - x) 2 x \end{matrix}$
$K_{P} = 0.5 P_{0}$
$Let, initial pressure of N_{2} O_{4} be P$
$At equilibrium (t = t e q), (p - x) + 2 x = 3 P_{0}$
$p + x = 3 P_{0} - \dots - (i)$
$And, K_{p} = \frac{{(P_{N O_{2}})}_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} \Rightarrow \frac{(2 x)^{2}}{p - x} = \frac{4 x^{2}}{p - x} = \frac{P_{0}}{2} \dots (ii)$
${From eq}^{n} (i) and (ii) n = \frac{1}{2} P_{0} and P = \frac{5}{2} P_{0}$
$\begin{matrix} When volume is halved, pressure will be doubled, due to Boyle's Law (P \propto \frac{1}{V}) \end{matrix}$
$so, \begin{matrix} P_{N_{2} O_{4}} & = 2 (P - x) = 2 \times 2 P_{0} = 4 P_{0} . P_{N O_{2}} & = 2 (2 x) = 2 P_{0} . \end{matrix}$
$Hence, new equilibrium pressure is 6 P_{0}$

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Kp for N2O2⇀2NO2 at equilibrium pressure of 3Po is 0.5Po. On halving the volume of container equilibrium is distributed. The new equilibrium pressure is :

$K_{p} f o r N_{2} O_{2} ⇀ {2 N O}_{2}$ at equilibrium pressure of ${3 P}_{o}$ is $0.5 P_{o}$ . On halving the volume of container equilibrium is distributed. The new equilibrium pressure is :