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9.2g of N₂O_4(g) is taken in 1lit vessel and heated. At equilibrium, 50% is dissociated. Equilibrium constant in (mol/lit) for the reaction N₂O_4(g) $⇌$ 2NO_2(g) is

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0.1

0.4

0.2

answer is B.

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Detailed Solution

\large {N_2}{O_4}\left( g \right) \rightleftharpoons 2N{O_2}\left( g \right)

Initial moles of N₂O₄

\large {\text{Degree of decomposition = }}\frac{{50}}{{100}} = 0.5

Number of moles of N₂O₄ decomposed = 0.1 x 0.5 = 0.05

1 mole of N₂O₄(g) → 2 moles of NO₂

0.05 mole of N₂O₄(g) → 'x' moles of NO₂

\large {K_C} = \frac{{\left[ {N{O_2}} \right]^2}}{{\left[ {{N_2}{O_4}} \right]}}

\large \boxed{{K_C} = \frac{\left ( 0.1 \right )^2}{\left ( 0.05 \right )}=0.2}

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