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) + 5 H_(2)O rightarrow CuSO_(4).5H_(2)O, Delta H = ?`
Given that
`CuSO _(4)(anhyd.) + aq. Rightarrow CuSO_(4)(aq.),` `Delta H_(1) = -66.4 kJ` ......(i)
`CuSO_(4).5H_(2)O + aq. rightarrow CuSO_(4) aq,` `Delta H_(2) = + 11.7 kJ` ....(ii)
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) = -78.1 kJ`