The equilibrium constant for the reaction
`SO_(3)(g) hArr SO_(2)(g)+(1)/(2)O_(2)(g)`
is `K_(e )=4.9xx10^(-2)`. The value of `K_(e )` for the reaction
`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` will be

Options

A. `9.8xx10^(-2)`
B. `4.9xx10^(-2)`
C. `416`
D. `2.40xx10^(-3)`

Answer

C

Explanation

`SO_(3)(g) hArr SO_(2)(g)+(1)/(2)O_(2)(g)`
`K_(e )=([SO_(2)][O_(2)]^(1//2))/([SO_(3)])=4.9xx10^(-2)`.
On taking the squate of the abov reaction
`([SO_(2)]^(2)[O_(2)])/([SO_(3)]^(2))=24.01xx10^(-4)`
now `K_(c) " for " 2SO_(2)(g)+O_(2)(g)hArr 2SO_(3)`
`=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=(1)/(24.01xx10^(-4))=416`