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The equilibirum pressure for the reactio...

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

A

`pi^(2)//4`

B

`pi//6`

C

`pi^(2)//16`

D

`pi//16`

Text Solution

Verified by Experts

The correct Answer is:
C
`underset("(Equilibrium pressure)")(K_(p)) = ("[Partial pressure of product]"^(x))/("[Partial pressure of reactant]"^(y)) = ((p)_("product")^(x))/((p)_("reactant")^(y))`
(x and y are powers over gaseous products and reactants respectively)
`therefore K_(p) = ((p)_("product")^(x))/((p)_("reactant")^(y)) = (P)_("Product")^(2) = ((pi)/(4))^(2) = (pi^(2))/(16) "atm"^(2)`
Therefore, `K_(p)` for given reaction
(in `"atm"^(2)`) `= (pi^(2))/(16)`
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