Calculate the enthalpy of hydration of anhydrous copper sulphate (CuSO4) into hydrated copper sulphate (CuSO4.5H2O). Given that the enthalpies of solution of anhydrous copper sulphate and hydrated copper sulphate are -66.5 and +11.7 kJ mol–1 respectively.
We are given
(i) CuSO4(s) + aq → CuSO4(aq); ΔsoiH = −66.5 kJ mol−1
(ii) CuSO45H2O(s) + aq → CuSO4(aq); ΔsoiH = +11.7 kJ mol−1
We aim at CuSO4(s) + 5H2O(l) → CuSO45H2O(s); ΔhydH = ?
Equation (i) can be written in two steps as:
(iii) CuSO4(s) + 5H2O(l) → CuSO45H2O(s); ΔH = ΔH1 kJ mol−1
(iv) CuSO45H2O(s) + aq → CuSO4(aq); ΔH = ΔH2 kJ mol−1
According to Hess's law, ΔH1 + ΔH2 = −66.5 kJ mol−1
Further, equation (ii) and (iv) are same;
∴ ΔH2 = +11.7 kJ mol−1
Putting this value above, we get ΔH1 + 11.7 = −66.5 or ΔH1 = −66.5 − 11.7 kJ = −78.2 kJ mol−1
Thus, equation (iii) may be written as
CuSO4(s) + 5H2O(l) → CuSO45H2O(s); ΔhydH = −78.2 kJ mol−1
This is what we aimed at. Hence, the required value of the enthalpy of hydration is ΔhydH = −78.2 kJ mol−1