Calculate the enthalpy change on freezing 1.0 mol of water at 10.0^(@)C to ice at - 10.0 ^(@)C, Delta_(fus) H= 6.03 kJ mol^(-1) mol^(-1) at 0^(@)C C_(p) [H_(2)O(l)]= 75.3J mol^(-1) K^(-1) C_(p)[ H_(2)O(s)] = 36.8 J mol^(-1) K^(-1)

Calculate the enthalpy change on freezing 1.0 mol of water at 10.0^(@)

1.	Calculate the enthalpy change on freezing 1.0 mol of water at 10.0^(@)C to ice at - 10.0 ^(@)C, Delta_(fus) H= 6.03 kJ mol^(-1) mol^(-1) at 0^(@)C C_(p) [H_(2)O(l)]= 75.3J mol^(-1) K^(-1) C_(p)[ H_(2)O(s)] = 36.8 J mol^(-1) K^(-1)
Answer» Solution :Total `DeltaH = ` ( 1 mol water at `10^(@)C rarr ` 1 mol of waterat `0^(@)C) + (` 1 mol water at `0^(@)C rarr 1 `mol ice at `0^(@)C )+(` 1 mol ice at `0^(@) Crarr `1 mol ice at `- 10^(@)C) ``(DeltaT = T_(2) - T_(1))` `=C_(p) [H_(2)O(l) ]xx DeltaT + DeltaH _("freezing") +C_(p)[H_(2)O (s) ] xx DeltaT ` `= ( 75.3J K^(-1) mol^(-1) ) ( 0- 10) K( - 6.03 kJ mol^(-1)) + ( 36.8 J K^(-1) mol^(-1) ) ( - 10 K) ( DeltaH_("freezing" ) =-DeltaH_("fusion") )` `=-753 J mol^(-1) - 603 kJ mol^(-1) - 368 J mol^(-1)` `= - 0.753 kJ mol^(-1) - 6.03 kJ mol^(-1) -0.368kJ mol^(-1) kJ mol^(-1) =-7.151kJ mol^(-1)` Note `:` Directly also, as in each STEP , heat is evolved,each step will have a NEGATIVE SIGN with `DeltaH`.