Consider the following equilibrium:
`SO_(3) rarr SO_(2)+O_(2)`
`8.0 g` of `SO_(3)` are put in a container at `600^(@)C`. The equilibrium pressure and density are `1.8` atm and `1.6 g L^(-1)`, respectively
a. Find the value of `K_(p)`.
b. Also find the moles of helium that is to be added at equilibrium to double the pressure at constant temperature.

Explanation

Weight of `SO_(3)` taken `=8.0 g`
molecular weight of `SO_(3)=80 g//"mol"`
`:.` mole of `SO_(3)=8/80=0.1 "mol"`
`{:(SO_(3),hArr,SO_(2),+,1//2O_(2)),(0.1-x,,x,,x//2):}`
`PV=nRT` or `PM=dRT`
Total moles `=0.1+x//2`
`rArr M=(dRT)/(P)=(1.6xx0.0821xx873)/(1.8)=63.6`
`n_(1)M_(1)=n_(2)M_(2), n_(1)/n_(2)=M_(2)/M_(1)`
`rArr (0.1)/(0.1+x//2)=M_(av)/M_(SO_(3))=63.6/80rArr x=0.05`
`:. K_(p)=(((x//2)/(0.1+x//2)P)^(1//2)((xP)/(0.1+x//2)))/(((0.1-x)/(0.1+x//2)P))=0.61`
`(P=1.8 "atm")`
b. `n_(i)=0.1 x//2`
To double pressure by adding inert gas at constant `T` double moles at equilibrium `rArr n_(f)=2(0.1+x//2)`
`rArr n_(He)` added `=0.1+x//2=0.1+(0.05//2)=0.125`