At a given temperature and a total pressure of `1.0` atm for the homogenous gaseous reaction
`N_(2)O_(4)hArr2NO_(2)(g),`
the partial pressure of `NO_(2)` is `0.5 atm`.
a. Find the value of `K_(p)`.
b. If the volume of the vessel is decreased to half of its original volume, at constant temperature, what are the partial pressure of the components of the equilibrium mixture?

Explanation

For equilibrium system, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the total pressure is 1.0 atm.
`rArr` The total pressure `=P_(N_(2)O_(4))+P_(NO_(2))=1.0` atm
`rArr R_(N_(2)O_(4))=0.5"atm and "P_(NO_(2))=0.5` atm
(i) `K_(p)=((P_(NO_(2)))^(2))/(P_(N_(2)O_(4)))=(0.5^(2))/(0.5)=0.5` atm
(ii) As volume is decreased to half its original volume, equilibrium is disturbed and the new initial conditions for the re-establishment of new equilibrium are:
`P_(N_(2)O_(4))=1.0` atm and `P_(NO_(2))=1.0` atm
According to Le Chatelier principle, when volume is decreased, the system moves in that direction where there is decrease in number of moles. Hence, the system (here) will move in reverse direction, as there is a decrease in mole `(Deltan_(g)=2-1=1)` i.e, the `NO_(2)` will be converted to `N_(2)O_(4)`.
Let the decrease in pressure of `NO_(2)` be x atm.

`rArr K_(p)=((1-x)^(2))/((1+x//2))=0.5 rArr 4x^(2)-9x+2=0 rArr x=2 or 0.25`
(`x ne 2` as initial pressure of `NO_(2)=1.0` atm)
`rArr x =0.25`
`P_(N_(2)O_(4))=1+(x)/(2)=1.125` atm and `P_(NO_(2))=1-x=0.75` atm