The equilibrium pressure for the reaction ` MSO_(4).2H_(2)O (s) hArr MSO_(4) (s) + 2H_(2)O (g) " is " pi //4 "atm at 400 K . The " K_(p) " for the given reaction ( in " atm^(2) ) ` is

Options

A. `pi^2//4`
B. `pi//6`
C. `pi^2//16`
D. `pi//16`

Answer

C

Explanation

The given reaction is :
`MSO_(4) . 2H_2O_((s)) hArr MSO_(4(s)) + 2H_2O_((g))`
As the pressure is due to `H_2O_((g))` only hence,
`K_p = (p_(H_2O))^2 =(pi/4)^2=pi^2/16"atm"^2`