For each of the given changes, state whether the final enthalpy is greater or less than the initial enthalpy:
(1) `H_(2)O (s) to H_(2)O (l)`
(2) `H_(2)O(g)toH_(2)O(l)`.

Explanation

Change (1) is a melting process. So, it is an endothermic process. Hence, in this process, `DeltaH=+ve`.
`DeltaH=H[H_(2)O(l)]-H[H_(2)O(s)]`. As `DeltaH gt0, H[H_(2)O(l)]gtH[H_(2)O(s)]`, indicating greater enthalpy for the final state than the initial state.
Change (2) is a condensation process and so, it is an exothermic process. hence in this process `DeltaH=-ve`.
`DeltaH=H[H_(2)O(l)]-H[H_(2)O(g)]`
Since `DeltaH lt 0, H[H_(2)O(l)]ltH[H_(2)O(g)]`. thus enthalpy of the final stae is less than that of the initial state.