An equilibrium mixture at 300K contains `N_2O_4 and NO_2` at 0.28 and 1.1 atm pressure respectively. If the volume of the container is doubled, calculate the new equilibrium pressure of the two gases.

Explanation

`N_2O_4 Leftrightarrow 2NO_2`
`"0.28 atm 1.1 atm at equilibrium"`
`K_(p)=((1.1)^(2))/(0.28)=4.32`
Further when the volume of the container is doubled, the pressure of `N_2O_4 and NO_2` shall be temporarily halved till some new equilibrium is attained.
`"0.14 at 0.55 atm when vol. is halved"`
`N_2O_4 Leftrightarrow 2NO_2`
`(0.14-x)atm " "(0.55+2x)"atm .... at the new eqb"`
`K_(p)=((0.55+2x)^2)/(0.14-x)=4.32 ("temp. remaining same")`
x=0.045
`p_(N_2O_4)=0.14-x=0.14 0.045=0.095 atm`
`and p_(NO_2)=0.55+2x=0.55 +2 xx 0.045 =0.64 atm`