Calculate the enthalpy of hydration of anhydrouscopper sulphate ( CuSO_(4)) into hydrated copper sulphate (CuSO_(4). 5H_(2)O). Given that the enthalpies of solutionsof anhydrous copper sulphate and hydrated copper sulphate are - 66.5 and + 11.7 kJ mol^(-1) respectively.

Calculate the enthalpy of hydration of anhydrouscopper sulphate ( CuSO

1.	Calculate the enthalpy of hydration of anhydrouscopper sulphate ( CuSO_(4)) into hydrated copper sulphate (CuSO_(4). 5H_(2)O). Given that the enthalpies of solutionsof anhydrous copper sulphate and hydrated copper sulphate are - 66.5 and + 11.7 kJ mol^(-1) respectively.
Answer» Solution :We are given (i) `CuSO_(4)(s) + aq rarr CuSO_(4)(aq), Delta_(sol)H= - 66.5 kJ mol^(-1)` (ii) `CuSO_(4) . 5H_(2)O(s) +aq rarr CuSO_(4)(aq), Delta_(sol) H= + 11.7 kJ mol^(-1)` we aim at`CuSO_(4) (s) + 5H_(2)O(l) rarr CuSO_(4).5H_(2)O(s), Delta _(HYD) H= ?` Eqaution (i) can be written in two steps as`:` `(ii) CuSO_(4) (s) +5H_(2)O(l) rarr CuSO_(4).5H_(2)O(s) , Delta H= Delta H_(1) kJ mol^(-1)` (iv)`CuSO_(4). 5H_(2)O(s)+ aq rarr CuSO_(4)(aq) , Delta H = DeltaH_(2) kJ mol^(-1)` According to Hess's law, `Delta H_(1) + Delta H_92) = - 66.5 kJ mol^(-1)` Further, equations (ii) and (iv) are same `:. Delta H_(2) = + 11.7 k J mol^(-1)` Putting this value above, we get `Delta H_(1) + 11.7 = - 66.5 ` or `Delta H_(1) = - 66.5- 11.7 kJ = - 78.2 kJ mol^(-1)` Thus, equation (iii) may be written as `CuSO_(4)(s) + 5H_(2)O(l) rarr CuSO_(4). 5H_(2)O(s), Delta _(hyd) H- 78.2kJ mol^(-1)` This is what we AIMED at . Hence, the required value of the ENTHALPY of hydration is`Delta _(hyd) H = - 78.2k J mol^(-1)`.