An equilibrium mixture at `300 K` contains `N_(2)O_(4)` and `NO_(2)` at `0.28` and `1.1 atm`, respectively. If the volume of container is doubles, calculate the new equilibrium pressure of two gases.

Explanation

`N_(2)O_(4) rarr 2NO_(2)`
`K_(p) = (P'_(NO_(2)))/(P'_(N)(2)O_(4)) = (1.1)^(2)/0.28 = 4.32 atm`
If volume of container is doubled, i.e., pressure becomes half, the reaction will also procced in the direction where the reaction shows an increase in mole. i.e., decomposition of `N_(2)O_(4)` is favoured.`N_(2)O_(4) rarr 2NO_(2)`
New pressure at equilibrium ` [0.28/2-p][1.1/2 + 2p]`
Where, `N_(2)O_(4)` equivalent to pressure P is used up in doing so.
Again, `K_(p) = [(1.1/2)+2p]^(2)/[(0.28/2)-P] = 4.32 implies P = 0.045`
`P_(N_(2)O_(4))` at new equilibrium = 0.14 - 0.045 = 0.095 atm
`P_(NO_(2))` at new equilibrium = `0.55 + 2xx0.045 = 0.64 atm`