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)))^(2))/(P_(N_(2)O_(4)))=((1.1)^(2))/(0.28)=4.32 atm`
Now if volume of container is doubled i.e., pressure decrease and will become half, the reaction will proceed 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),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])/([0.4-P])=4.32`
`P=0.045`
`:.` Now `P_(N_(2)O_(4))=0.14-0.045=0.095 atm`
`P_(NO_(2))` at new eqm. `=0.55+2xx0.045`
`P_(NO_(2))=0.64 atm`