Calculate the enthalpy change of freezing of `1.0` mol of water at `10^(@)C` to ice at `-10^(@)C, Delta_(fus)H=6.03 kJ mol^(-1)` at `0^(@)C`.
`C_(P)[H_(2)O(l)]=75.3 J mol^(-1) K^(-1)`
`C_(P)[H_(2)O(s)]=36.8 J mol^(-1) K^(-1)`

Answer

`DeltaH=-7.151 kJ mol^(-1)`

Explanation

Total enthalpy change involved in the transfromation is the sum of the following changes :
(a) Energy change involved in the transformation of 1 mol of water at `1^(@)C` to 1 mol of water at `0^(@)`
(b) Energy change involved in the transformation of 1 mol of water at `0^(@)C` to mol of ice at `0^(@)C`
(c) Energy change involved in the transformation of 1 mol of ice at `0^(@)C` to 1 mol of ice at `10^(@)C`
Total`DeltaH=C_(p)[H_(2)OCI]DeltaT+DeltaH_("freezing")+C_(p)[H_(2)O_(s)]DeltaT`
`=(75.3 J "mol"^(-1)H^(-1))(0-10)H+(-6.03xx10^(3) J "mol"^-1)+(36.8 J "mol"^(-1) H^(-1))(-10-0)K`
`-753 J "mol"^(-1)-6030 J "mol"^(-1)-368 J "mol"^(-1)`
`=-7151 J "mol"^(-1)`
`=-7.151 kJ "mol"^(-1)`
Hence, the enthalpy change involved in the transformation is `-7.151 kJ "mol"^(-1)`