The equilibrium constant of the reaction,
`SO_(3)(g) hArr SO_(2)(g)+1//2 O_(2)(g)`
is `0.15` at `900 K`. Calculate the equilibrium constant for
`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`

Answer

C, D

Explanation

For `SO_(3)(g) hArr SO_(2)(g)+1/2 O_(2)`
`K_(C_(1))=([SO_(2)][O_(2)]^(1//2))/([SO_(3)])=0.15` …(i)
For `2SO_(2)+O_(2) hArr 2SO_(3)`
`K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])` …(ii)
By reversing equation (i), `1/K_(C_(1))=([SO_(3)])/([SO_(2)][O_(2)]^(1//2))`
On making square,
`(1/K_(C_(1)))^(2)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=K_(c)` by equation (ii)
`:. K_(c)=[1/0.15]^(2)=44.44`