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At a given temperature and a total press...

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?

Text Solution

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For equilibrium system , `N_2O_4(g) hArr 2NO_2(g)` , the total pressure is 1.0 atm
`rArr` The total pressure `=P_(N_2O_4)+P_(NO_2)=1`
`rArr P_(N_2O_4)=0.5 "atm" and P_(NO_2) =0.5 "atm"`
(i) `K_p=((P_(NO_2))^2)/(P_(N_2O_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 new equilibrium are
`P_(N_2O_4)=1.0 "atm" and P_(NO_2)=1.0 "atm"`
According to lo Chatelier.s 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 decreasse in mole (`Deltan=2-1=1`) i.e `NO_2` will be converted to `N_2O_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 (xne2 "as initial pressure" =1.0) rArr x=0.25`
`P_(N_2O_4)=1+x/2=1.125 "atm" and P_(NO_2)=1-x=0.75 "atm"`
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