When 1 mole of anhydrous `CuSO_(4)` is dissolved in excess of water, -66.4 kJ heat is evolved. When one mole of `CuSO_(4).5H_(2)O` is dissolved in water, the heat change is +11.7 kJ. Calculate enthalpy of hydration of `CuSO_(4)` (anhydrous).

Explanation

We have to calculate enthalpy of reaction for
`CuSO_(4) + 5H_(2)O rarr CuSO_(4).5H_(2)O, Delta H =`?
Given that
`CuSO_(4) ("anhyd.") + aq rarr CuSO_(4)(aq.), Delta H_(1)` = -66.4 kJ
`CuSO_(4).5H_(2)O + aq rarr CuSO_(4)aq., Delta H_(2)` = +11.7 kJ
On the basis of two equations
`Delta H = Delta H_(1) - Delta H_(2)`
= -66.4 kJ - (+11.7 kJ)
= -78.1 kJ
The enthalpy of hydration of `CuSO_(4)` = -78.41 kJ