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

Consider the reaction,
`{:(,2SO_(3),hArr,2SO_(2),+,O_(2)(K_(p)=(1)/(900)atm)),("Initial pressure",1,,0,,2),("Pressures left at ",(1-x),,x,,2+(x)/(2)):}`
equilibrium
`:. K_(p)=((P'_(SO_(2)))^(2)(P'_(O_(2))))/((P'_(SO_(3)))^(2))`
`(1)/(900)=(x^(2)(2+(x)/(2)))/((1-x)^(2))`
`:. ` Since `K_(p)` of this reaction is small and thus, `xlt1`
`(1)/(900)=(x^(2)(2))/((1-x)^(2))` (`:' 2+(x)/(2)=2`)
`=(2x^(2))/((1-x)^(2))`
`:. (1)/(30)=(sqrt(2)x)/((1-x))`
`:. x=0.0236`
At equilibrium,
`:. P'_(SO_(3))=1-x=1-0.236=0.9764 atm`
`P'_(SO_(2))=x=0.0236 atm`
`P'_(O_(2))=2+(x)/(2)=2.0118 atm`