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

Answer

D

Explanation

`SO_(2)(g) + (1)/(2)O_(2)(g) rarr SO_(3)(g) K_(p) = 4xx10^(-3)`
`SO_(3) rarr SO_(2)(g) + (1)/(2) O_(2)(g)K'_(p) = (1)/(K_p)`
`K'_(p) = ((1)/(4xx10^(-3)))`
`2SO_(3) rarr 2SO_(2) + O_(2)(g)`
`K''_(p) = (K'_(p))_(2) = [(1)/(4xx10^(-3))]^(2) = [(1000000)/(4)]`
`= 6250 = 625xx10^(2) = 6.25xx10^(4)` atm