At `0^(@)C`, ice and water are in equilibrium and `Delta H^(@)` = 6.00 kJ/mol for the process
`H_(2)O(s) hArr H_(2)O(I)`
Value of `Delta S^(@)` for the conversion of ice to liquid water is
`{:((1),10.15 JK^(-1)mol^(-1),(2),17.25 JK^(-1)mol^(-1)),((3),21.98 JK^(-1)mol^(-1),(4),30.50 JK^(-1)mol^(-1)):}`

Explanation

`H_(2)O(s) hArr H_(2)O(I)" "Delta H^(@)="6.00 kJ mol"^(-1)`
`Delta S^(@)=(Delta H^(@))/(T)=(6.00 xx 10^(3)"J mol"^(-1))/(273 K)`
= 21.98 `JK^(-1)mol^(-1)`
`therefore "Ans"(3)`