18.4 g `N_(2)O_(4)` was placed in 1 L vessel at 400 K and allowed to attain the following equilibrium `N_(2)O_(4)(g) hArr 2NO_(2)(g)`. IF the total pressure at equilibrium was 10.64 bar, approximate `K_(p)` is (R = 0.083 L bar `K^(-1) mol^(-1)`) (Assume `N_(2)O_(4), NO_(2)` as ideal gases)

Options

A. `57.20`
B. `24.24`
C. `14.30`
D. `6.64`

Answer

D

Explanation

From ideal gas equation,
pV = nRT
`p_(N_(2)O_(4)) xx 1 = (18.4)/(92) xx 0.083 xx 400`
`p_(N_(2)O_(4)) = 6.64` bar
For the equilibrium reaction,
`{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial pressure",p,,0),("At equilibrium","p-pi",,"2 pi"):}`
`p_(T) = p - p_(i) + 2p_(i)`
`p_(T) = p + p_(i) rArr p_(i) = p_(T) - p`
`p_(i) = 10.6- 6.64`
`= 4.00` bar
At equilibrium,
`:. p_(N_(2)O_(4)) = p - p_(i) = 6.64 - 4`
`= 2.64` bar
`p_(NO_(2)) = 2p_(i) = 2 xx 4`
= 8 bar
`:. K_(p) = ([p_(NO_(2))]^(2))/([p_(N_(2)O_(4))])`
`= ([8]^(2))/(2.64) = 24.29`