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),hArr,2NO_(2)),("Pressure at equilibrium",0.28,,1.1):}`
`:. K_(p)=((p_(NO_(2)))/p_(N_(2)O_(4)))^(2)=((1.1)^(2))/(0.28)=4.32 "atm"`
Now if the volume of container is doubled, i.e., pressure decreases and will become half, the reaction will proceed in the direction where the reaction shows an increase in moles i.e., decomposition of `N_(2)O_(4)` is favoured.

`{:(,,N_(2)O_(4),hArr,2NO_(2)),("New pressure at equilibrium",,(0.28/2-P),,(1.1/2+2P)):}`
where reactant `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])=([0.55+2P]^(2))/([.14-P])=4.32`
`P=0.045`
`:.` Now `P_(N_(2)O_(4))=0.14-0.045=0.095 "atm"`
`P_(NO_(2))` at new equilibrium=`0.55+2xx0.045`
`:. P_(NO_(2))=0.64 "atm"`