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

From the unit of `K_p` it is obvious that `K_p=900` is for the equilibrium
`overset(1)(2SO_(3)) Leftrightarrow overset(0)(2SO_(2))+overset(2)(O_(2))" Initial pressure"`
`"(1-2x) 2x (2+x) Eqb. Pressure"`
where 2x is the partial pressure of `SO_2` at equilibrium.
Thus, `K_p=(p_(SO_(2))^(2) (p_(O_(2))))/(p_(SO_(3))^(2))=((2x)^2 (2+x))/((1-2x)^(2))=900`
On solving we get `x=0.475`
`p_(SO_(3))=1-2x=1-2 xx 0.475 =0.05atm`
`p_(SO_(2))=2x=2xx 0.475 =0.95atm`
and `p_(O_(2))=2+x=2+0.475=2.475atm`.