The equilibrium constant of the reaction ,
`SO_(3(g))hArrSO_(2(g))+(1)/(2)O_(2(g))`,
is `0.20 "mole"^(1//2)litre^(-1//2)` at `1000 K`. Calculate equilibrium constant for
`2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`

Explanation

For `SO_(3(g))hArrSO_(2(g))+(1)/(2)O_(2(g))`
`K_(c_(1))=([SO_(2)][O_(2)]^(1//2))/([SO_(3)])=0.20` …(`1`)
For `2SO_(2)+O_(2)hArr2SO_(3)`
`K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])` …(`2`)
By reversing eq. (`1`),
`(1)/(K_(c_(1)))=([SO_(3)])/([SO_(2)][O_(2)]^(1//2))`
Squaring both sides by eq. (`2`),
`((1)/(K_(c_(1))))=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=K_(c)`
`:. K_(c)=[(1)/(0.20)]^(2)=25mol^(-1)litre^(+1)`