The equilibrium constant of the reaction `SO_(2)(g) + 1//2O_(2)(g)hArrSO_(3)(g)` is `4xx10^(-3)atm^(-1//2)`. The equilibrium constant of the reaction `2SO_(3)(g)hArr2SO_(2)(g) + O_(2)` would be:

Options

A. `250atm`
B. `4xx10^(3) atm`
C. `0.25xx10^(4)atm`
D. `6.25xx10^(4)atm`

Answer

D

Explanation

`SO_(2)+(1)/(2)O_(2)(g)hArrSO_(3)(g) K_(p)=4xx10^(-3)`
`SO_(3)hArrSO_(2)(g)+(1)/(2)O_(2)(g) K'_(p)(1)/(KP)`
` K'_(p)=((1)/(4xx10^(-3)))`
`2SO_(3)hArr2SO_(2)+O_(2)(g)`
`K"_(p)=(K'_(p))^(2)=[(1)/(4xx10^(-3))]^(2)=[(1000)/(4)]^(2)=6250=625xx10^(2) 6.25xx10^(4)`atm.