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

Now : `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g) " " K_(p)=900` atm
`rArr 2SO_(3)(g) hArr 2SO_(2)(g) +O_(2)(g)" "K_(p)=1//900` atm
and initial pressure of `SO_(3)=` atm and `O_(2)=2` atm.
Let x be the decrease in partial pressure of `SO_(3)` at equilibrium.

`rArr K_(p)=((P_(SO_(2)))^(2)xxP_(O_(2)))/((P_(SO_(3)))^(2)) rArr K_(p)=(x^(2)xx(2+(x)/(2)))/((1-x)^(2))=(1)/(900)`
`rArr (x^(2)(4+x))/(2(1-x)^(2))=(1)/(900)`
Assume that x is very small (as `K_(p) lt lt 1` and `O_(2)` is already present at the time of dissociation).
`rArr (4+x)~4 and (1-x)~1`
`rArr (x^(2)(4))/(2(1)^(2))=(1)/(900)rArr x=(1)/(30sqrt(2))=0.023`
`rArr P_(SO_(2))=x=0.023` atm
and `P_(SO_(3))=1-x=0.977` atm
and `P_(O_(2))=2+(x)/(2)=2.0115` atm