The equilibrium constant of the reaction at `25^(@)C`
`CuSO_(4).5H_(2)O(s)iffCuSO_(4).3H_(2)O(s)+3H_(2)O(g)`
is `1.084xx10^(-4)atm^(2)`. Find out under what conditions of relative humidity. `CuSO_(4).5H_(2)O` will start loosing its water of crystallization according to above reaction. (Vapour pressure of water at `25^(@)C "is"24` `mm` of `Hg`).

Explanation

`K_(P)=(P_(H_(2)o))^(2) salpha P_(H_(2)o=sqrt(1.084xx10^(-4))=1.041xx10^(-2) atmapprox8mm "of" Hg`
ᴏ If in a room, pressure of waer is greater than `8 mm` of Hg then `CuSO_(4).3H_(2)O` will absorb water from air and will form `CuSO_(4).3H_(2)O` & will keep absorbing until partial pressure of `H_(2)O` becomes `8 mm` of Hg.
ᴏ If `P_(H_(2o)lt8 mm` of Hg then `CuSO_(4).5H_(2)O` will loose water of crystalization and reaction will move in forward direction.
i.e. if relative humidiy`lt(8)/(24)xx33.33%`
then `CuSO_(4).5H_(2)O`will loose water of crystalization.