The equilibrium constant `K_(p)` of the reaction: `2SO_(2)+O_(2) hArr 2SO_(3)` is `900 atm^(-1)` at `800 K`. A mixture constaining `SO_(3)` and `O_(2)` having initial pressure of 1 atm and 2 atm respectively, is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at `800 K` at equilibrium.

Explanation

`{:(,2SO_3 hArr , 2SO_2 + ,O_2 , K_p=1/900"atm"),("Initial pressure", 1,0,2),("Pressure left at equilibrium",(1-x), x, 2+x/2):}`
`therefore K_p=((P'_(s(O_2)))(P'_(O_2)))/((P'_(SO_3)^2) rArr 1/900=(x^2(2+x/2))/((1-x)^2)`
`therefore` Since, Kp of this reaction is small and thus, x <<< 1, `therefore x^3` is negligible
`therefore 1/900=(x^2(2))/((1-x)^2)=(2x^2)/((1-x)^2), 1/30 = (sqrt2x)/((1-x)), x=0.0236`
`P'_(SO_3)=1-x=1-0.0236` = 0.9764 atm `P'_(SO_2)` = x = 0.0236 atm , `P'_(O_2) = 2+x/2`= 2.0118 atm