An equilibrium mixture at 300K contains N _ 2 O _ 4 and NO _ 2 at...

{N}_{2}{O}_{4}\left(g\right) ⇌ 2{NO}_{2}\left(g\right)$</p><p>Pressure at equilibrium<br />${K}_{p} = {p}_{{NO}_{2}}^{2}{p}_{{N}_{2}{O}_{4}} = \dfrac{{\left(1 . 1\right)}^{2}}{0 . 28} = 4 . 32atm$</p><p>If volume of the container is doubled, the pressure will reduced to half</p><p>${N}_{2}{O}_{4} ⇌ 2{N}_{2}O$</p><p>New pressure,</p><p>\left[\dfrac{0 . 028}{2} - p\right]\left[\dfrac{1 . 1}{2} + 2p\right]$

${K}_{p} = \dfrac{{\left[\dfrac{1 . 1}{2} + 2p\right]}^{2}}{\left[\dfrac{0 . 28}{2} - p\right]} = 4 . 32$

On solving, we get

$p = 0 . 045$

$∴ {p}_{{N}_{2}{O}_{4}} = 0 . 14 - 0 . 045 = 0 . 095atm$

${p}_{{NO}_{2}} = 0 . 55 + \left(2 \times 0 . 045\right) = 0 . 64atm$