At `25^(@)C` and `1 atm` pressure, the partial pressure in equilibrium mixture of `N_(2)O_(4)` and `NO_(2)`, are `0.7` and `0.3 atm`, respectively. Calculate the partial pressures of these gases when they are in equilibrium at `25^(@)C` and a total pressure of `10 atm`.

Explanation

`{:(,,N_(2)O_(4),hArr,2NO_(2)),("Pressure at equilibrium",,0.7,,0.3):}`
`:. K_(p)=((p_(NO_(2)))^(2))/(p_(N_(2)O_(4)))=(0.3xx0.3)/(0.7)=0.1286 "atm"`
Now assume decomposition at `1.0 "atm"` pressure
`{:(,,N_(2)O_(4),hArr,2NO_(2)),("Initial "moles"",,1,,0),(""moles" at equilibrium",,(1-x),,2x):}`
`K_(p)=((n_(NO_(2)))^(2))/(n_(N_(2)O_(4)))xx(P/(Sigman))^(Deltan)=((2x)^(2))/((1-x))xx[10/((1+x))]^(1)`
or `0.1286=(4x^(2)xx10)/((1-x)^(2)) :. x=0.0565`
`P'_(NO_(2))=(2x)/((1+x))xxP=(2xx0.0565xx10)/((1+0.0565))=1.07 "atm"`
`P'_(N_(2)O_(2))=[(1-0.0565)/(1+0.0565)]xx10=8.93 "atm"`