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The equilibrium constant for the reactio...

The equilibrium constant for the reaction `CaSO_(4).5H_(2)O(s)hArr CaSO_(4).3H_(2)O(s)+2H_(2)O(g)` is equal to

A

`([CaSO_(4).3H_(2)O][H_(2)O]^(2))/([CaSO_(4).5H_(2)O])`

B

`([CaSO_(4).3H_(2)O])/([CaSO_(4).5H_(2)O])`

C

`[H_(2)O]^(2)`

D

`[H_(2)O]`

Text Solution

Verified by Experts

The correct Answer is:
C
Concentration of solids is taken to be unity.
Thus `[CaSO_(4).5H_(2)O(s)]`
`[CaSO_(4).3H_(2)O(s)]=1`
and `K=[H_(2)O]^(2)`
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